# Does alternating test show divergence?

My book states the alternating tests' convergence requirements.

However, my book doesnt point out, if $a_n$ fails one of the convergence requirements, is it true that is diverges? Such as the limit as $n \to \infty$

If

$$\lim_{n \to \infty} a_n \ne 0$$

then does $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}a_n$ diverge Or does the test all together fail?

• If $a_n$ has a non-zero limit then the alternating series sum will certainly not converge to a finite number Dec 18, 2014 at 12:18

A necessary condition for a series to converge is that $$\lim_{n \to \infty} a_n = 0$$

This has nothing to do with the alternating series test. If one of the other hypothesis fails, then one cannot conclude divergence.

• So the test itself fails, and we have to try something else/ Dec 18, 2014 at 12:21
• @Amad27 you are correct :)
– Ant
Dec 18, 2014 at 12:22
• The alternating test says $\lim_{n\to \infty} a_n = 0$ is a requirement for convergence, above you have said it has nothing to do with the test? Dec 18, 2014 at 12:43
• @Amad27 $\lim_{n \to \infty} a_n = 0$ is a requirement for every series, not just alternating series. I mean it really is very general; of course it is a requirement for alternating series, because is a requirement for every series
– Ant
Dec 18, 2014 at 12:46

The following alternating series fails the alternating series test with flying colors $$\sum_{n=1}^{\infty} (-1)^n\dfrac{\cos^2\left(n\right)}{n^2}.$$

This series is however is absolutely convergent, which is easy to show.

In addition, you can take many absolutely convergent series which do pass the test and create one which does not:

Given $\ \sum\limits_{n=1}^{\infty} (-1)^n b_n\$ where $\sum\limits_{n=1}^{\infty} b_n$ converges, and $b_n < b_{n-1}$ for all $n$. Make a new series by defining $a_{2n-1} = b_{2n}$ and $a_{2n}=b_{2n-1}$. Then $\{a_n\}$ is not decreasing, as for all $n$, $a_{2n-1}=b_{2n}<b_{2n-1}=a_{2n}$. However, because the original series is absolutely convergent

$$\sum\limits_{n=1}^{\infty} (-1)^n b_n = \sum\limits_{n=1}^{\infty} (-1)^n a_n$$