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.
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