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Ratio Test Quick Breakdown

used to prove convergence

yes

used to prove divergence

yes

can be inconclusive

yes

if \(L=\infty\) then \(L>1\) and the series diverges

\(a_n\) terms can be positive or negative or both

requires the use of limits at infinity

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Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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free ideas to save on books

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 best way to get a feel for this is to build a set of sheets 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.
Do you know why? Click here for the answer.

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

Okay, time for some practice problems.

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - 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 Problems

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

Problem Statement

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

Final Answer

The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n!}{2^n} \right] } }\) diverges by the Ratio Test.

Problem Statement

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

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. That includes anything that you don't understand or is not clear on this site.

Final Answer

The series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n!}{2^n} \right] } }\) diverges by the Ratio Test.

close solution

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

Problem Statement

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

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n+2}{2n+9} \right] } }\) diverges by the divergence test.

Problem Statement

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

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.

Final Answer

\(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{n+2}{2n+9} \right] } }\) diverges by the divergence test.

close solution

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

Problem Statement

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

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n} \right] } }\) diverges by the ratio test.

Problem Statement

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

Solution

207 solution video

video by MIT OCW

Final Answer

The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n}{n \cdot 3^n} \right] } }\) diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{10}}{10^n} \right] } }\)

Solution

208 solution video

video by MIT OCW

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{2^n+5}{3^n} \right] } }\)

Solution

209 solution video

video by Dr Chris Tisdell

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{3^k}{k^2} \right] } }\)

Solution

211 solution video

video by Krista King Math

Final Answer

The series diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\)

Final Answer

The series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\) converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\)

Solution

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

212 solution video

video by Dr Chris Tisdell

212 solution video

video by Krista King Math

Final Answer

The series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{2^k}{k!} \right] } }\) converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ n(5/4)^n \right] } }\)

Solution

213 solution video

video by MIP4U

Final Answer

The series diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^{3n}}{(2n)!} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n 2^{3n}}{(2n)!} \right] } }\)

Solution

214 solution video

video by MIP4U

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^3}{(\ln 3)^n} \right] } }\)

Solution

215 solution video

video by Educator.com

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^2}{2^n} \right] } }\)

Solution

216 solution video

video by PatrickJMT

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(-3)^n}{4^{n-1}} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n(-3)^n}{4^{n-1}} \right] } }\)

Solution

217 solution video

video by PatrickJMT

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (n^2)2^n}{n!} \right] } }\)

Final Answer

The series converges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^{n+1} (n^2)2^n}{n!} \right] } }\)

Solution

219 solution video

video by PatrickJMT

Final Answer

The series converges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{7^n}{e^n} \right] } }\)

Solution

220 solution video

video by PatrickJMT

Final Answer

The series diverges by the ratio test.

close solution

Intermediate Problems

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{\sum_{n=2}^{\infty}{\left[\frac{n^2 2^n}{5^n}\right]}}\)

Final Answer

The series converges by the Ratio Test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{\sum_{n=2}^{\infty}{\left[\frac{n^2 2^n}{5^n}\right]}}\)

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.

\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{a_{n+1}}{a_n} \right|} }\)

\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2 2^{n+1}}{5^{n+1}} \frac{5^n}{n^2 2^n} \right|} }\)

\(\displaystyle{ \lim_{n \to \infty}{\left| \frac{(n+1)^2}{n^2} \frac{2^{n+1}}{2^n} \frac{5^n}{5^{n+1}} \right|} }\)

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

Final Answer

The series converges by the Ratio Test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{3n^2}{(2n-1)!}\right]}}\)

Final Answer

The series converges absolutely by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{\sum_{n=0}^{\infty}{\left[\frac{3n^2}{(2n-1)!}\right]}}\)

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.

Final Answer

The series converges absolutely by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(-3)^{n+1}n} \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{9^n}{(-3)^{n+1}n} \right] } }\)

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.

Final Answer

The series diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(-1)^n}{n^2+1} \right] } }\)

Final Answer

The series converges by the alternating series test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{3(-1)^n}{n^2+1} \right] } }\)

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.

Final Answer

The series converges by the alternating series test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(2n)!}{n!n!} \right] } }\)

Solution

210 solution video

video by Dr Chris Tisdell

Final Answer

The series diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\)

Solution

218 solution video

video by PatrickJMT

Final Answer

The series diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }\)

Final Answer

The series diverges by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{4^n(n!)^2}{(n+2)!} \right] } }\)

Solution

221 solution video

video by PatrickJMT

Final Answer

The series diverges by the ratio test.

close solution

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^n}{n!} } }\)

Solution

1872 solution video

video by Dr Chris Tisdell

close solution

Advanced Problems

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

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(-5)^n}{4^{2n+1}(n+1)} \right] } }\)

Final Answer

The series converges absolutely by the ratio test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \left[ \frac{(-5)^n}{4^{2n+1}(n+1)} \right] } }\)

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 \(5/16 < 1\) 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.

Final Answer

The series converges absolutely by the ratio test.

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