# Calculus/Divergence Test

## Divergence Test

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 Definition

If ${\displaystyle \displaystyle {\lim _{n\to \infty }{a_{n}}\neq 0}}$, then ${\displaystyle \displaystyle {\sum _{n=1}^{\infty }{a_{n}}}}$ diverges.

### Divergence Test Quick Notes

 used to prove convergence no used to prove divergence yes can be inconclusive yes

## How To Use The Divergence Test

To use the divergence test, just take the limit ${\displaystyle \displaystyle {\lim _{n\to \infty }{a_{n}}}}$. 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 ${\displaystyle \displaystyle {\sum _{n=1}^{\infty }{a_{n}}}}$, if ${\displaystyle \displaystyle {\lim _{n\to \infty }{a_{n}}\neq 0}}$, 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 ${\displaystyle \displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n}}}}$ ${\displaystyle \displaystyle {\lim _{n\to \infty }{\frac {1}{n}}=0}}$ diverges p-series with ${\displaystyle p=1}$ B ${\displaystyle \displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}}$ ${\displaystyle \displaystyle {\lim _{n\to \infty }{\frac {1}{n^{2}}}=0}}$ converges p-series with ${\displaystyle p=2}$

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.

## Examples

Determine whether or not these series are divergent or if the divergence test is inconclusive.

### Example 1

${\displaystyle \sum _{n=0}^{\infty }{\frac {n+1}{n}}}$

Because ${\displaystyle \lim _{n\rightarrow \infty }{\frac {n+1}{n}}=1}$, the limit is not zero and so the series is divergent by the divergence test.

### Example 2

${\displaystyle \sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}}$

Because ${\displaystyle \lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}={\frac {1}{3}}}$ the limit is not zero and so the series is divergent by the divergence test.

### Example 3

${\displaystyle \sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{n^{3}}}}$

Because ${\displaystyle \lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{n^{3}}}=0}$ the limit is zero and so the test is inconclusive. Further analysis is needed to determine whether or not the series converges or diverges.