## 17Calculus Infinite Series - Divergence Test

##### 17Calculus

The divergence test is the easiest infinite series test to use but students can get tripped up by using it incorrectly. On this page, we explain how to use it and how to avoid one of the most common pitfalls associated with this test.
The Divergence Test is also called the nth-Term Test.

Divergence Test

If $$\displaystyle{\lim_{n \to \infty}{a_n} \neq 0}$$, then $$\displaystyle{\sum_{n=1}^{\infty}{a_n}}$$ diverges.

### Mathematical Logic Form

$\lim_{n \to \infty}{a_n} \neq 0 ~ \to ~ \sum_{n=1}^{\infty}{a_n} \text{ diverges}$ 17calculus Mathematical Logic

### Alternate Form

The Divergence Test can also be stated in this alternate form.

If $$\displaystyle{\sum_{n=1}^{\infty}{a_n}}$$ converges, then $$\displaystyle{\lim_{n \to \infty}{a_n} = 0}$$.

Using Mathematical Logic this can be written $$\displaystyle{ \sum_{n=1}^{\infty}{a_n} \text{ converges } ~ \to ~ \lim_{n \to \infty}{a_n} = 0 }$$, which is the contrapositive of the form above.

This alternate form is not very useful for computation. So we use the theorem above for all discussion and practice problems here at 17calculus.

How To Use The Divergence Test

To use the divergence test, just take the limit $$\displaystyle{\lim_{n \to \infty}{a_n} }$$. If this limit turns out to be non-zero, the series diverges and you are done. If the limit is equal to zero, then the test is inconclusive and says nothing about the series. It may converge or it may diverge. You need to use another test to determine convergence or divergence.

Things To Watch Out For

The Divergence Test seems to be pretty straight-forward. Basically, this test says that, for a series $$\displaystyle{ \sum_{n=1}^{\infty}{a_n}}$$, if $$\displaystyle{ \lim_{n \to \infty}{a_n} \neq 0 },$$ then the series diverges. Pretty simple, eh? However, there is a trap that many students fall into.

Trap - - There is a tendency by almost every student to use this theorem to prove convergence. The statement says nothing about convergence. Let us give you a couple of examples that demonstrate what we mean. Let's compare the convergence or divergence of these two very similar series. (If you don't know about p-series yet, just take our word for the convergence/divergence conclusions. You will understand this soon enough.)

 A. $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n}}}$$ $$\displaystyle{\lim_{n\to\infty}{\frac{1}{n}} = 0}$$ diverges p-series with $$p=1$$ B. $$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2}}}$$ $$\displaystyle{\lim_{n\to\infty}{\frac{1}{n^2}} = 0}$$ converges p-series with $$p=2$$

It is important to notice that in both cases the limit of the terms goes to zero. However, series A diverges while series B converges. (See the Infinite Series - Integral Test page for a video showing a proof of the convergence/divergence of these two series.)

So you can see that just because the limit goes to zero, this does not guarantee the series will converge. The only time you can apply this theorem is when the limit does not go to zero. This guarantees divergence. When the limit does go to zero, you still don't know if the series converges or diverges. You need to use another test to determine convergence.

Below is the divergence test row from the infinite series table. Notice that the convergence line says that this test cannot be used. Think about this thoroughly and completely so that you get your head around it.

Divergence Test
$$\displaystyle{ \sum_{n=1}^{\infty}{a_n} }$$

convergence:

this test cannot be used

divergence:

$$\displaystyle{ \lim_{n \to \infty}{ a_n \neq 0} }$$

### njc314 - Here is a quick video that explains this test. [1min-26secs]

video by njc314

Okay, so you are now ready for some practice problems. Once you go through those, the next logical step is the p-series.

Practice

Unless otherwise instructed, determine whether these series converge or diverge.

Basic

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

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2} } }$$ converges or diverges.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2} } }$$ diverges by the Divergence Test.

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2} } }$$ converges or diverges.

Solution

Let's try the Divergence Test.
$$\displaystyle{ \lim_{n \to \infty}{n/2} = \frac{\infty}{2} = \infty }$$
Since the limit is not zero, the series diverges by the Divergence Test.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{2} } }$$ diverges by the Divergence Test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }$$

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }$$ converges or diverges.

Convergence/divergence cannot be determined by the divergence test. The series converges by the p-series test.

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5n}{n^3} } }$$ converges or diverges.

Solution

Starting with the divergence test, we need to determine the limit $$\displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} } }$$.
There are two main ways to do this.
1. The first, and easiest way, is to use algebra.
2. The second way is to use L'HÃ´pital's Rule.
We will use algebra by multiplying the numerator and denominator by $$1/n$$.
$$\begin{array}{rcl} \displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} }} & = & \displaystyle{ \lim_{n \to \infty}{ \frac{5n}{n^3} \frac{1/n}{1/n} }} \\ & = & \displaystyle{ \lim_{n \to \infty}{ \frac{5}{n^2}}} = 0 \end{array}$$
Since the limit is zero, we cannot say anything about whether this series converges or diverges using only the divergence test.
If you have not studied the p-series yet, skip the next section and jump to the notes.
p-series test - By canceling the $$n$$ in the numerator with one in the denominator (which we do since the canceling operation does not change the domain), we have $$a_n = 5/n^2$$. This is a p-series with $$p=2>1$$. So this series converges by the p-series test.
Notes
1. Even though the divergence test cannot be used to determine convergence or divergence, we didn't just stop working the problem. The problem statement said to determine convergence or divergence. So stopping after applying the divergence test would not be the completion of the problem. We actually needed to determine convergence or divergence. So once the first test failed, we moved on to another test until one of the tests determined convergence or divergence.
2. If you have not studied the p-series test yet, do not despair. Just make sure you understand that you cannot determine convergence/divergence from the divergence test when the limit is zero and that you can't just stop working the problem when the divergence test doesn't work.

Convergence/divergence cannot be determined by the divergence test. The series converges by the p-series test.

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

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+3n+1}{2n^2+4} } }$$ converges or diverges.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+3n+1}{2n^2+4} } }$$ diverges by the Divergence Test.

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+3n+1}{2n^2+4} } }$$ converges or diverges.

Solution

### PatrickJMT - 130 video solution

video by PatrickJMT

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+3n+1}{2n^2+4} } }$$ diverges by the Divergence Test.

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

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+2}{n^2+1} } }$$ converges or diverges.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+2}{n^2+1} } }$$ diverges by the Divergence Test.

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+2}{n^2+1} } }$$ converges or diverges.

Solution

### PatrickJMT - 132 video solution

video by PatrickJMT

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2+2}{n^2+1} } }$$ diverges by the Divergence Test.

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Intermediate

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

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \cos\left[ \frac{1}{n^2} \right] } }$$ converges or diverges.

Solution

### blackpenredpen - 3199 video solution

video by blackpenredpen

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }$$

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }$$ converges or diverges.

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }$$ diverges by the Divergence Test.

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }$$ converges or diverges.

Solution

### PatrickJMT - 131 video solution

video by PatrickJMT

The series $$\displaystyle{ \sum_{n=1}^{\infty}{ \ln(\cos n) } }$$ diverges by the Divergence Test.

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

Problem Statement

Determine whether the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{n~\cos(n\pi)}{2n-1}}}$$ converges or diverges.

The series diverges by the divergence test.

Problem Statement

Determine whether the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{n~\cos(n\pi)}{2n-1}}}$$ converges or diverges.

Solution

$$\displaystyle{ a_n = \frac{n~\cos(n\pi)}{2n-1} }$$

$$\displaystyle{\lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } = \lim_{n \to \infty}{ \frac{\cos(n\pi)}{2-1/n} } }$$

In this limit, we know that $$\displaystyle{ \lim_{n \to \infty}{1/n} = 0 }$$, leaving $$2$$ in the denominator. In the numerator, $$\cos(n\pi)$$ oscillates between $$-1$$ and $$1$$. So we are left with the series oscillating between $$-1/2$$ and $$1/2$$ and therefore, the limit $$\displaystyle{\lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } \neq 0 }$$
So, by the divergence test, the series diverges.
The divergence test is the best test for this series. For a discussion on using other tests, see the practice problem page dedicated to this problem.

The series diverges by the divergence test.

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$$\displaystyle{ \sum_{n=1}^{\infty}{ \sin(n\pi/2) } }$$

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \sin(n\pi/2) } }$$ converges or diverges.

The series $$\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}$$ diverges by the Divergence Test.

Problem Statement

Determine whether the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \sin(n\pi/2) } }$$ converges or diverges.

Solution

In order to determine what the limit $$\displaystyle{\lim_{n\to\infty}{\sin(\pi n/2)}}$$ does, you need to think about the sine function. Basically, the sine function takes angles and maps them to the range $$-1 \to 1$$. So as $$n \to \infty$$ the sine function oscillates between -1 and 1. It never settles at a specific value and, more importantly for us here, it never settles or approaches zero.
Conclusion: Since $$\displaystyle{\lim_{n\to\infty}{\sin(\pi n/2)}\neq 0}$$, this series diverges by the Divergence Test.

The series $$\displaystyle{\sum_{n=1}^{\infty}{\sin(n\pi/2)}}$$ diverges by the Divergence Test.

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