## 17Calculus Infinite Series - Ratio Test

##### 17Calculus

The Ratio Test is probably the most important test and the test you will use the most as you are learning infinite series. It is used A LOT in power series. I believe it is the most powerful test of all and, if you look at the Infinite Series Table, you will see that it is listed first in Group 3. So I suggest you master it from the start. It's not hard, and if your algebra skills are strong, you might even find it fun to use. Also, the more familiar you are with it and the more practice problems you work, the sooner you will start to be able to look at a series and see almost right away if the Ratio Test will tell you what you need to know. Cool, eh?

Ratio Test

Let $$\sum{a_n}$$ be a series with nonzero terms and let $$\displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} = L}$$

Three cases are possible depending on the value of L.
$$L < 1$$: The series converges absolutely.
$$L = 1$$: The Ratio Test is inconclusive.
$$L > 1$$: The series diverges.

When To Use The Ratio Test

The ratio test is best used when you have certain elements in the sum. The way to get a feel for this is to build a set of tables containing examples of tests that work as you are working practice problems. This is one technique listed in the infinite series study techniques section. This is an extremely powerful technique that will help you really understand infinite series.
Here is a list of things to watch for.
1. Sums that include factorials.
2. Sums with exponents containing n.

How To Use The Ratio Test

In general, the idea is to set up the ratio $$\displaystyle{\lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} = L}$$ and evaluate it.
In detail, you need to determine what $$a_n$$ is and then build $$a_{n+1}$$, set up the fraction, combine like terms and then take the limit of each term. Setting up the limit and combining like terms are the easy parts. The challenge comes in taking the limit.
Key - It is important to remember to use the absolute value signs unless you are absolutely convinced that the term will always be positive. This is critical to practice up front since, once you get to Taylor Series, you can't and don't want to drop the absolute value signs. They are critical to the result. It is never wrong to include them and, as you work more problems, you will get a feel for when you need them and when you don't. In the practice problems and examples, we will use them unless we explicitly state that they are not needed. Some instructors are less rigid about this than others. As always, check with your instructor to see what they require.

Things To Notice

1. If you get $$\infty$$ for the limit, this indicates divergence since it fits the case where the limit is greater than one. Notice that the theorem says nothing about the limit needing to be finite.

2. In the fraction that you are taking the limit of, the $$n+1$$ term is in the numerator and the $$n$$ term is in the denominator. In order for the ratio test to work, they must appear exactly like this.

3. If you get L=1, you cannot say anything about convergence or divergence of the series. You need to use another test. Sometimes, a comparison test (either direct or limit) will be the best next step.

Okay, it's time for some videos. This first video clip is a great overview of the ratio test. Notice that he doesn't use absolute value signs, so he requires that the terms be positive.

### Dr Chris Tisdell - Series, Comparison + Ratio Tests [1min-50secs]

video by Dr Chris Tisdell

The beginning of this next video has a good discussion about the ratio test. Then the instructor shows two examples when the ratio test is inconclusive to emphasize that a series may converge or diverge when the ratio test is inconclusive.

### MIT OCW - Ratio Test for Convergence [9min-19secs]

video by MIT OCW

Ratio Test Proof

Here are a couple of proofs of the Ratio Test.

This first video contains a rather long and involved proof of the ratio test. It uses a comparison test.

### Linda Green - Proof of the Ratio Test

video by Linda Green

Here is a second proof presented in 3 separate videos.

### slcmath@pc - The Ratio Test - Proof of Part (a)

video by slcmath@pc

### slcmath@pc - The Ratio Test - Proof of Part (b)

video by slcmath@pc

### slcmath@pc - The Ratio Test - Proof of Part (c)

video by slcmath@pc

You do not need to watch these proofs in order to use and understand the Ratio Test. We are including them here for those of you who are interested.

Before You Start Working The Practice Problems

For these practice problems, take a few minutes and scan them. You notice there are a LOT of very different ones (there are even more than we have included here - we plan to continue adding more in the future). The key to solving infinite series problems is to find patterns so that you can quickly narrow down the techniques that might work to about 2 or 3. This is especially true with problems on which the ratio test works. Read more about how to solve infinite series problems on the infinite series study techniques page.
Once you have done that you are ready for these practice problems.

Practice

Unless otherwise instructed, determine the convergence or divergence of the following series, using the ratio test, if possible. [These instructions imply that if the ratio test fails ($$L = 1$$), you need to use another test to prove convergence or divergence.]

Basic

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{5^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3190 video solution

video by blackpenredpen

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$$\displaystyle{ \sum_{k=1}^{\infty}{ k (2/3)^k } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ k (2/3)^k } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3196 video solution

video by blackpenredpen

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$$\displaystyle{ \sum_{k=1}^{\infty}{ k^2 e^{-k} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ k^2 e^{-k} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3197 video solution

video by blackpenredpen

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$$\displaystyle{ \sum_{n=0}^{\infty}{ \left[\frac{n!}{2^n}\right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[\frac{n!}{2^n}\right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the Ratio Test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[\frac{n!}{2^n}\right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

Whenever you have a factorial, the ratio test will often work. So, let's try it.
$$\displaystyle{ a_n = \frac{n!}{2^n} }$$ and $$\displaystyle{ a_{n+1} = \frac{(n+1)!}{2^{n+1}} }$$
$$\displaystyle{\lim_{n \to \infty}{ \left| \frac{a_{n+1}}{a_n} \right|} = \lim_{n \to \infty}{ \left| \frac{(n+1)!}{2^{n+1}} \div \frac{n!}{2^n} \right| } = \lim_{n \to \infty}{ \left| \frac{(n+1)!}{2^{n+1}} \cdot \frac{2^n}{n!} \right| } }$$
Now combine like terms.
$$\displaystyle{ \frac{2^n}{2^{n+1}} = \frac{2^n}{2(2^n)} = 1/2 }$$
$$\displaystyle{\frac{(n+1)!}{n!} = \frac{1\cdot2\cdot3\cdot4 . . . n\cdot(n+1)}{1\cdot2\cdot3\cdot4 . . . n} = n+1 }$$
So our limit is now $$\displaystyle{ \lim_{n \to \infty}{ \left| \frac{n+1}{2} \right|} = \frac{\infty}{2} = \infty > 1 \to }$$ the series diverges.
Note - We could have left off the absolute value signs all the way through this problem. However, to follow the theorem exactly, we need them unless we explicitly state that the ratio is positive. It is a common practice (although not always good) to leave them off without explaining, which teachers may do quite often. If you don't understand something, always ask.

The series diverges by the Ratio Test.

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$$\displaystyle{ \sum_{n=0}^{\infty}{ \left[\frac{n+2}{2n+9}\right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[\frac{n+2}{2n+9}\right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the divergence test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[\frac{n+2}{2n+9}\right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

$$\displaystyle{ a_n = \frac{n+2}{2n+9} }$$
$$\displaystyle{ a_{n+1} = \frac{n+3}{2(n+1)+9} = \frac{n+3}{2n+11} }$$
Now we can set up the limit and evaluate it.

 $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{n+3}{2n+11} \frac{2n+9}{n+2} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+3)(2n+9)}{(2n+11)(n+2)} \frac{1/n^2}{1/n^2} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(1+3/n)(2+9/n)}{(2+11/n)(1+2/n)} \right|} = \frac{2}{2} = 1 }$$

So the ratio test failed. When the limit is equal to one, we cannot tell if the series converges or diverges.
However, the divergence test yields
$$\displaystyle{ \lim_{n \to \infty}{\frac{n+2}{2n+9}} = \frac{1}{2} \neq 0 }$$
Therefore, the series diverges by the divergence test.
Notes
- Notice that we did not just stop after the ratio test failed. We needed to use another test to complete the problem, since we were told to determine convergence or divergence in the problem statement.
- Once you get more experience with infinite series, you will realize that the ratio test is not the best test to use here. Whenever you have polynomials, one in the numerator and one in the denominator, check out the ratio of the highest power terms. If you get a constant, that is a good indication that maybe the series diverges and the divergence test may be the quickest way to determine divergence.

The series diverges by the divergence test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n } \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n } \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n } \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### MIT OCW - 207 video solution

video by MIT OCW

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### MIT OCW - 208 video solution

video by MIT OCW

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### Dr Chris Tisdell - 209 video solution

video by Dr Chris Tisdell

The series converges by the ratio test.

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$$\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### Krista King Math - 211 video solution

video by Krista King Math

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

We have included two solutions to this problem by two different instructors.

### Dr Chris Tisdell - 212 video solution

video by Dr Chris Tisdell

### Krista King Math - 212 video solution

video by Krista King Math

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3^n}{n!} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3^n}{n!} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### The Organic Chemistry Tutor - 3178 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### MIP4U - 213 video solution

video by MIP4U

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^{3n}}{(2n)!} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^{3n}}{(2n)!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^{3n}}{(2n)!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### MIP4U - 214 video solution

video by MIP4U

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### Educator.com - 215 video solution

video by Educator.com

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{4^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{4^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### The Organic Chemistry Tutor - 3179 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### TYEducational - 3187 video solution

video by TYEducational

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{e^n} } }$$

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{e^n} } }$$ converges or diverges using the Ratio Test.

Solution

### slcmath@pc - 3257 video solution

video by slcmath@pc

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### PatrickJMT - 216 video solution

video by PatrickJMT

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n^3}{2^n} } }$$

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^n n^3}{2^n} } }$$ converges or diverges using the Ratio Test.

Solution

### slcmath@pc - 3258 video solution

video by slcmath@pc

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(-3)^n}{4^{n-1}} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(-3)^n}{4^{n-1}} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(-3)^n}{4^{n-1}} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### PatrickJMT - 217 video solution

video by PatrickJMT

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^n}{(-2)^n n!} } }$$

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^n}{(-2)^n n!} } }$$ converges or diverges using the Ratio Test.

Solution

### slcmath@pc - 3259 video solution

video by slcmath@pc

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (n^2)2^n}{n!} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (n^2)2^n}{n!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (n^2)2^n}{n!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### PatrickJMT - 219 video solution

video by PatrickJMT

The series converges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### PatrickJMT - 220 video solution

video by PatrickJMT

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{n!}{2 \cdot 5 \cdot 8 \cdots (3n+2) } } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{n!}{2 \cdot 5 \cdot 8 \cdots (3n+2) } } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### TYEducational - 3185 video solution

video by TYEducational

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$$\displaystyle{ \sum_{k=1}^{\infty}{ \frac{1 \cdot 4 \cdot 7 \cdots (3k-2)}{2 \cdot 4 \cdot 6 \cdots (2k)} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \frac{1 \cdot 4 \cdot 7 \cdots (3k-2)}{2 \cdot 4 \cdot 6 \cdots (2k)} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### TYEducational - 3186 video solution

video by TYEducational

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Intermediate

$$\displaystyle{ \sum_{k=1}^{\infty}{ \frac{2^k k!}{(k+2)!} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \frac{2^k k!}{(k+2)!} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### TYEducational - 3184 video solution

video by TYEducational

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(n+2)}{\sqrt{n}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(n+2)}{\sqrt{n}} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the divergence test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(n+2)}{\sqrt{n}} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

The ratio test is inconclusive for this series. The video does not go on to use another test to determine convergence or divergence, so he never completes the problem. Since this is an alternating series, let's start there.

 $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n+1}(n+2)}{\sqrt{n}} } = \sum_{n=1}^{\infty}{ (-1)^{n+1} a_n } }$$ where $$\displaystyle{ a_n = \frac{(n+2)}{\sqrt{n}} }$$ According to the alternating series test, we have two conditions that must be met for convergence. Remember that the alternating series test cannot be used to prove divergence. If both conditions are fulfilled, then the series will converge. Condition 1: $$\displaystyle{ \lim_{n \to \infty}{ a_n } }$$ Condition 2: $$0 \lt a_{n+1} \lt a_n$$ for all $$n$$ greater than some $$N \gt 0$$ Let's start with condition 1. $$\displaystyle{ \lim_{n \to \infty}{ a_n } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \frac{n+2}{\sqrt{n}} } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \left[ \frac{n}{n^{1/2}} + \frac{2}{n^{1/2}} \right] } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \frac{n}{n^{1/2}} } + }$$ $$\displaystyle{ \lim_{n \to \infty}{ \frac{2}{n^{1/2}} } }$$ The second limit $$\displaystyle{ \lim_{n \to \infty}{ \frac{2}{n^{1/2}} } = 0}$$ The first limit $$\displaystyle{ \lim_{n \to \infty}{ \frac{n}{n^{1/2}} } = }$$ $$\displaystyle{ \lim_{n \to \infty}{ n^{1/2} } = \infty }$$ So the first condition fails. However, we CANNOT say that the series diverges by the alternating series test. We need to notice that this condition is also the divergence test. So we can say that the series diverges BY THE DIVERGENCE TEST.

### The Organic Chemistry Tutor - 3182 video solution

The series diverges by the divergence test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-3)^{n-1}}{\sqrt{n}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-3)^{n-1}}{\sqrt{n}} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### TYEducational - 3183 video solution

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$$\displaystyle{\sum_{n=2}^{\infty}{\left[\frac{n^2 2^n}{5^n}\right]}}$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{\sum_{n=2}^{\infty}{\left[\frac{n^2 2^n}{5^n}\right]}}$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the Ratio Test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{\sum_{n=2}^{\infty}{\left[\frac{n^2 2^n}{5^n}\right]}}$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

Since we have $$n$$'s in the exponents, we might be tempted to use the Root Test. However, the Root Test is best used when the base also has an $$n$$ in it. So, let's use the Ratio Test.

 $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2 2^{n+1}}{5^{n+1}} \frac{5^n}{n^2 2^n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2}{n^2} \frac{2^{n+1}}{2^n} \frac{5^n}{5^{n+1}} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \left(\frac{n+1}{n}\right) \frac{2}{5} \right|} = 2/5 }$$

The series converges by the Ratio Test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3^{n+2}\cdot n^2}{4^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3^{n+2}\cdot n^2}{4^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### The Organic Chemistry Tutor - 3180 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2 2^{n-1}}{(-5)^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2 2^{n-1}}{(-5)^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3192 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{e^{n^2}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{e^{n^2}} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3193 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3^n n^2}{n!} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3^n n^2}{n!} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3194 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^k k!}{(k+2)!} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^k k!}{(k+2)!} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3195 video solution

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$$\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{3n^2}{(2n-1)!}\right]}}$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{3n^2}{(2n-1)!}\right]}}$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges absolutely by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{3n^2}{(2n-1)!}\right]}}$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

Since we have a factorial, let's try the ratio test.
$$\displaystyle{ a_n = \frac{3n^2}{(2n-1)!} }$$
$$\displaystyle{ a_{n+1} = \frac{3(n+1)^2}{(2(n+1)-1)!} = \frac{3(n+1)^2}{(2n+1)!} }$$
Set up the limit and evaluate it.

 $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{3(n+1)^2}{(2n+1)!} \frac{(2n-1)!}{3n^2} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2}{n^2} \frac{(2n-1)!}{(2n+1)!} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2}{n^2} \right. } }$$ $$\displaystyle{ \left. \frac{(2n-1)!}{(2n+1)(2n)(2n-1)!} \right| }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2}{n^2} \right. } }$$ $$\displaystyle{ \left. \frac{1}{(2n+1)(2n)} \frac{1/n^4}{1/n^4} \right| }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{1/n + 1/n^2}{1(2+1/n)(2)} \right|} = \frac{0}{4} = 0 < 1 }$$

Since the limit $$0 < 1$$ the series converges absolutely.

The series converges absolutely by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(-3)^{n+1}n} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(-3)^{n+1}n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(-3)^{n+1}n} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

$$\displaystyle{ a_n = \frac{9^n}{(-3)^{n+1}n} }$$ Let's rewrite this a little bit to make the limit simpler.
$$\begin{array}{rcl} a_n & = & \displaystyle{ \frac{9^n}{(-3)^{n+1}n} } \\ & = & \displaystyle{ \frac{9^n}{(-1)^{n+1}3^{n+1}n} } \\ & = & \displaystyle{ \frac{(-1)^{n+1} 9^n}{3^{n+1}n} } \end{array}$$
In the above work, we can move the $$-1$$ term to the numerator since the fraction is the same whether the sign is in the numerator or denominator. This is the only time we can do to this. So, now that $$a_n$$ is simpler, let's write $$a_{n+1}$$.
$$\displaystyle{ a_{n+1} = \frac{(-1)^{n+2} 9^{n+1}}{3^{n+2}(n+1)} }$$
And, now we can calculate the limit.

 $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{9^{n+1}}{3^{n+2}(n+1)} \frac{3^{n+1}n}{9^n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{9^{n+1}}{9^n} \frac{3^{n+1}}{3^{n+2}} \frac{n}{n+1} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{9}{3} \frac{n}{n+1} \right|} = 3 > 1 }$$

Since the limit $$3 > 1$$ the series diverges.

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{k=0}^{\infty}{ \frac{k!}{3^{k^2}} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=0}^{\infty}{ \frac{k!}{3^{k^2}} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

While solving this problem, he incorrectly introduces n into the problem. His n's should be k's.
Also at the end of the problem he has the limit $$\displaystyle{ \lim_{k \to \infty}{ \left| \frac{(k+1)}{3 \cdot 9^k} \right| } }$$ and then just says this is equal to zero but does not show the steps. Here are the steps.

 $$\displaystyle{ \lim_{k \to \infty}{ \left| \frac{k+1}{3 \cdot 9^k} \right| } }$$ $$\displaystyle{ \frac{1}{3} \lim_{k \to \infty}{ \left| \frac{k+1}{9^k} \right| } }$$ $$\displaystyle{ \frac{1}{3} \lim_{k \to \infty}{ \left| \frac{k}{9^k} + \frac{1}{9^k} \right| } }$$ We can drop the absolute values since both factors will always be positive. $$\displaystyle{ \frac{1}{3} \lim_{k \to \infty}{ \frac{k}{9^k} } + \frac{1}{3} \lim_{k \to \infty}{ \frac{1}{9^k} } }$$ The second limit is zero. We will use L'Hopitals Rule to evaluate the first limit. But first, we will convert $$9^k$$ to base $$e$$. If we let $$y = 9^k \to \ln y = k \ln 9$$ $$\displaystyle{ e^{\ln y} = e^{k\ln 9} \to y = e^{k\ln 9} \to 9^k = e^{k \ln 9} }$$ In order to use L'Hopitals Rule, we need a continuous function. So rewrite the limit with the continuous variable $$x$$. $$\displaystyle{ \frac{1}{3} \lim_{x \to \infty}{ \frac{x}{e^{x\ln 9}} } }$$ Now apply L'Hopitals Rule. $$\displaystyle{ \frac{1}{3} \lim_{x \to \infty}{ \frac{1}{(\ln 9)e^{x\ln 9}} = 0 } }$$ So $$\displaystyle{ \lim_{k \to \infty}{ \left| \frac{k+1}{3 \cdot 9^k} \right| } = 0 }$$

### TYEducational - 3188 video solution

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$$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(-1)^n}{n^2+1} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(-1)^n}{n^2+1} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges by the alternating series test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(-1)^n}{n^2+1} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

$$\displaystyle{ a_n = \frac{3(-1)^n}{n^2+1} }$$
$$\displaystyle{ a_{n+1} = \frac{3(-1)^{n+1}}{(n+1)^2+1} }$$
Now we can set up and evaluate the limit.

 $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{3}{(n+1)^2+1} \frac{n^2+1}{3} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{n^2+1}{(n+1)^2+1} \frac{1/n^2}{1/n^2} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{1+1/n^2}{(1+1/n)^2+1/n^2} \right|} = \frac{1}{1} = 1 }$$

So the ratio test failed. When the limit is equal to one, we cannot tell if the series converges or diverges.
However, if your first inclination was to use the alternating series test, you were right. We show in a practice problem on the alternating series page that this series converges absolutely.

Notes:
- Notice that we did not just stop after the ratio test failed. We needed to use another test to complete the problem, since we were told to determine convergence or divergence in the problem statement.
- Once you get more experience with infinite series, you will realize that the ratio test is not the best test to use here. With an alternating series, it is usually best to start with the alternating series test.

The series converges by the alternating series test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### Dr Chris Tisdell - 210 video solution

video by Dr Chris Tisdell

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^{4n+1}}{n^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2^{4n+1}}{n^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### The Organic Chemistry Tutor - 3181 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### PatrickJMT - 218 video solution

video by PatrickJMT

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series diverges by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### PatrickJMT - 221 video solution

video by PatrickJMT

The series diverges by the ratio test.

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$$\displaystyle{ \sum_{k=1}^{\infty}{ \frac{3^k}{k^k} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{k=1}^{\infty}{ \frac{3^k}{k^k} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

At the end of the video, he has the limit $$\displaystyle{ \lim_{k \to \infty}{ \left| \frac{3}{k+1} \cdot \left( \frac{k}{k+1} \right)^k \right| } }$$ and he says that each piece goes to zero, giving $$0 \cdot 0$$. The first fraction obviously goes to zero but the second does not. However, the final answer is correct.
The limit of the second fraction is not trivial and it is solved on the L'Hopital's Rule page as practice problem 3198.

### TYEducational - 3189 video solution

video by TYEducational

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{n^n} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{n^n} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

### blackpenredpen - 3191 video solution

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^n}{n!} } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^n}{n!} } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

This problem is solved by two different instructors.

### Dr Chris Tisdell - 1872 video solution

video by Dr Chris Tisdell

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$$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(-5)^n}{4^{2n+1}(n+1)} \right] } }$$

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(-5)^n}{4^{2n+1}(n+1)} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

The series converges absolutely by the ratio test.

Problem Statement

Determine the convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(-5)^n}{4^{2n+1}(n+1)} \right] } }$$, using the ratio test if possible. [These instructions imply that if the ratio test fails (L = 1), you need to use another test to prove convergence or divergence.]

Solution

Let's try the ratio test.
$$\displaystyle{ a_n = \frac{(-5)^n}{4^{2n+1}(n+1)} }$$
Before we set up $$a_{n+1}$$, let's simplify this a bit.

 $$\displaystyle{ a_n = \frac{(-5)^n}{4^{2n+1}(n+1)} }$$ $$\displaystyle{ a_n = \frac{(-1)^n 5^n}{4 \cdot 4^{2n} (n+1)} }$$ $$\displaystyle{ a_n = \frac{(-1)^n 5^n}{4 (16)^n (n+1)} }$$ $$\displaystyle{ a_n = \frac{1}{4} \left( \frac{5}{16} \right)^n \frac{(-1)^n}{n+1} }$$

Okay, so this $$a_n$$ will be simpler to work with and, since $$1/4$$ is just a constant, it does not affect convergence or divergence and can be taken outside the sum. This leaves
$$\displaystyle{ a_n = \frac{ (-1)^n 5^n}{16^n (n+1)} ~~~ \to ~~~ a_{n+1} = \frac{ (-1)^{n+1} 5^{n+1}}{16^{n+1} (n+2)} }$$
So now we set up the limit and evaluate it.

 $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{5^{n+1}}{16^{n+1} (n+2)} \right. } }$$ $$\displaystyle{ \left. \frac{16^n (n+1)}{5^n} \right| }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{5^{n+1}}{5^n} \frac{16^n}{16^{n+1}} \frac{n+1}{n+2} \right|} }$$ $$\displaystyle{ \lim_{n \to \infty}{\left| \frac{5}{16} \frac{n+1}{n+2} \right|} }$$ $$\displaystyle{ \frac{5}{16} < 1 }$$

Since the limit 5/16 is always less than 1 the series converges absolutely.
Note: This is also an alternating series, so we could have used the alternating series test to determine that it converges absolutely.

The series converges absolutely by the ratio test.

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### ratio test 17calculus youtube playlist

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