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Divergence Test

Divergence Test

Sometimes proving that a series diverges can be quite a challenge! Using the Divergence Test, also called the \(n^{th}\) Term Test for Divergence, is a simple test you can do to see right away if a series diverges. This can save you considerable time in the long run.

Divergence Tests in Calculus

Many of the tests used for series will have a part that also talks about divergence.

For example, the Direct Comparison Test and the Limit Comparison Test both have a part that talks about convergence and another that talks about divergence. The Integral Test, Ratio Test, and Root Test do as well. Some series, such as the P-series, Geometric series, and Arithmetic series, have known conditions for when they diverge and converge. So when you are looking for divergence tests, be sure to look at Convergence tests as well.

Series Divergence Tests

Here you will see a test that is only good to tell if a series diverges. Consider the series

\[\sum_{n=1}^{\infty} a_n,\]

and call the partial sums for this series \(s_n\). Sometimes you can look at the limit of the sequence \({a_n}\) to tell if the series diverges. This is called the \(n^{th}\) term test for divergence.

\(n^{th}\) term test for divergence.

If

\[\lim\limits_{n\to \infty}a_n\]

does not exist, or if it does exist but is not equal to zero, then the series

\[\sum_{n=1}^{\infty}a_n\]

diverges.

What is the wrong way to use the test?

The most common mistake people make is to say that if the limit of the sequence is zero, then the series converges. Let's take a look at an example to show why that isn't true.

Can you use the \(n^{th}\) term test for divergence to say that if

\[\lim\limits_{n\to \infty}a_n=0\]

the series converges?

Solution

Let's take a look at two examples.

First look at the Harmonic series

\[\sum_{n=1}^{\infty}\frac{1}{n}.\]

For this series, we have

\[\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\frac{1}{n}=0,\]

but you know that the series diverges.

Next look at the P-series with \(p=2\),

\[\sum_{n=1}^{\infty}\frac{1}{n^2}.\]

If you look at the limit of the sequence of terms for this series, you get,

\[\begin{align}\lim\limits_{n\to\infty}a_n&=\lim\limits_{n\to \infty}\frac{1}{n^2}\\ &=0\end{align}\]

as well, but this series converges.

So in fact, if the limit is zero, the series might converge and it might diverge, you just can't tell.

Proof of the \(n^{th}\) Term Test for Divergence

Let's take a look at the \(n^{th}\) term test for divergence is true. Sometimes in math, you prove a statement like "if A is true then B is true", and sometimes it is easier to prove the contrapositive, which is "if B is false then A is false".

For the \(n^{th}\) term test for divergence, it is easier to show the contrapositive.

So what is the contrapositive for the \(n^{th}\) term test for divergence?

The statement B is "the series diverges", and saying that "B is false" is the same as saying "the series converges".

The statement A is "the limit of the sequence either doesn't exist or does exist and isn't zero", and "A is false" is the same as saying "the limit of the sequence is zero". That means we will look at the proof of:

If \(\sum\limits_{n=1}^{\infty}a_n\) converges then

\[\lim\limits_{n\to \infty}a_n=0.\]

To do this, you will need to look at the partial sums for the series. The sequence of partial sums is defined by \[s_n=\sum_{k=1}^{n}a_k.\]

The previous term in the sequence of partial sum would be \[s_{n-1}=\sum_{k=1}^{n-1}a_k\]

Subtracting them gives,

\[\begin{align} s_{n}-s_{n-1}&=\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n-1}a_{k}\\&=(a_{1}+a_{2}+\cdots+a_{n-1}+a_{n})-(a_{1}+a_{2}+\cdots+a_{n-1})\\&=a_{n}.\end{align}\]

You already know that the series converges, which implies that the sequence of partial sums converges as well, or in other words

\[\lim\limits_{n\to\infty}s_n=L\]

for some real number \(L\). Now taking the limit of the subtraction of partial sums,

\[\begin{align}\lim\limits_{n\to\infty}[s_n-s_{n-1}]&=\lim\limits_{n\to \infty}s_n-\lim\limits_{n\to\infty}s_{n-1}\\&=L-L\\&=0.\end{align}\]

But you also know that

\[\lim\limits_{n\to\infty}[s_{n}-s_{n-1}]=\lim\limits_{n\to\infty} a_n,\]

which means that

\[\lim\limits_{n\to\infty}a_n=0.\]

Examples Using the \(n^{th}\)Term Test for Divergence

Let's look at some examples of how to properly use the \(n^{th}\) term test for divergence.

What can you say about the convergence or divergence of the series

\[\sum_{n=1}^{\infty}\frac{2n+3}{7n-1}\]

using the \(n^{th}\)term test for divergence?

Solution

For this series, \[a_{n}=\frac{2n+3}{7n-1},\]

and

\[\begin{align}\lim\limits_{n\to\infty}a_n &=\lim\limits_{n\to\infty}\frac{2n+3}{7n-1} \\ &=\frac{2}{7}.\end{align}\]

So the sequence converges, but the limit isn't zero. Then by the \(n^{th}\) term test for divergence, the series diverges.

Let's take a look at another example.

What can you say about the convergence or divergence of the series \[\sum_{n=1}^{\infty}(-1)^n,\]

using the \(n^{th}\) term test for divergence?

Solution

For this series, \(a_{n}=(-1)^n\), and the limit of this sequence doesn't exist. So by \(n^{th}\) term test for divergence, the series diverges.

Integral Divergence Test

As mentioned already, the Integral Test has a part that talks about divergence. So for more information on divergence when using the Integral Test, see Integral Test.

Divergence Test - Key takeaways

  • Many tests can be used to tell if a series converges or diverges, such as the Integral Test or the Limit Comparison Test.
  • The \(n^{th}\) term test for divergence is a good first test to use on a series because it is a relatively simple check to do, and if the series turns out to be divergent you are done testing.
  • If \[\sum\limits_{n=1}^{\infty}a_{n}\] converges then \[\lim\limits_{n\to\infty}a_n=0.\]

  • \(n^{th}\) term test for divergence: If \[\lim\limits_{n\to\infty}a_{n}\]

    does not exist, or if it does exist but is not equal to zero, then the series \[\sum_{n=1}^{\infty} a_n\] diverges.

Frequently Asked Questions about Divergence Test

It is a way to look at the limit of the terms of a series to tell if it diverges.  

Look at the limit of the terms of the series.  If that limit is not zero then the series diverges.

The divergence test is a good one to use on any series as a basic check.  If the divergence test shows the series doesn't converge, then you are done testing.

By looking at the difference of the partial sums.

We say that a series is convergent if its partial sum converges to a finite limit. If the partial sum does not have a finite limit, or the limit does not exist, the series is divergent. 

Final Divergence Test Quiz

Question

What is a good first test to do to see if a series diverges or converges?

Show answer

Answer

The nth term test for divergence.  It is relatively simple to apply it to see if the series diverges.  If it diverges, you are done testing.

Show question

Question

Is the nth term test for divergence the only divergence test?

Show answer

Answer

No, there are lots of tests that also look at divergence in addition to convergence, e.g. the Integral Test, the Ratio Test, and the Root Test.  Not to mention the Direct Comparison Test and the Limit Comparison Test.

Show question

Question

What is the wrong way to use the \(n^{th}\) term divergence test?

Show answer

Answer

The most common mistake people make is to say that if the limit of the sequence is zero, then the series converges.

Show question

Question

If 

\[\lim\limits_{n\to \infty}a_n\] 

does not exist what can you say about convergence of 

\[\sum_{n=1}^{\infty}a_n?\]

Show answer

Answer

By the \(n^{th}\) term test for divergence, you can say the series diverges.

Show question

Question

If 

\[\lim\limits_{n\to \infty}a_n \ne 0\] what can you say about convergence of the series 

\[\sum_{n=1}^{\infty}a_n?\]


Show answer

Answer

By the \(n^{th}\) term test for divergence, you can say the series diverges.

Show question

Question

If 

\[\lim\limits_{n\to \infty}a_n = 0\] what do you know about convergence of the series 

\[\sum_{n=1}^{\infty}a_n ?\]


Show answer

Answer

Nothing.  That isn't enough information to tell if the series converges or diverges.

Show question

Question

Is it true that if  \(\sum\limits_{n=1}^{\infty}a_n\) converges then

\[\lim\limits_{n\to \infty}a_n=0?\]


Show answer

Answer

Yes.  This is the contrapositive of the \(n^{th}\) term test for divergence.

Show question

Question

 Is it true that if \(\sum\limits_{n=1}^{\infty}a_n\) diverges then

\[\lim\limits_{n\to \infty}a_n=0?\]


Show answer

Answer

No, just knowing the series diverges doesn't tell you anything about the limit of the sequence.

Show question

Question

Why do you use the \(n^{th}\) term test on a new series as a first test?

Show answer

Answer

Because it is relatively simple to use and can save you a lot of work checking the various convergence tests.

Show question

Question

Other than the \(n^{th}\) term test, what is another test used to check for divergence?

Show answer

Answer

The Integral Test.

Show question

Question

What is the contrapositive of the statement "if A is true then B is true"?

Show answer

Answer

The contrapositive is "if B is false then A is false" 

Show question

Question

Give an example of a series where 

\[\lim\limits_{n\to \infty}a_n=0\]

but the corresponding series converges.

Show answer

Answer

The P-series with \(p=2\)

Show question

Question

Give an example of a series where 

\[\lim\limits_{n\to \infty}a_n=0\]

but the corresponding series diverges.

Show answer

Answer

The Harmonic Series.

Show question

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