Alternating divergent series I know that a necessary condition for a series to converge is that the general term $(a_n)$ goes to zero, but is it possible for an alternating divergent series to have a general term that goes to zero? If so, can you give an example.
My reasoning is that since $(a_n)\rightarrow 0$ then $\sum a_n$ converges. Then it can not be such series.
 A: It is indeed possible to have an alternating series with a general term going to zero diverge. Consider the sequence
$\frac{1}{1}, -\frac{1}{1}, \frac{1}{2}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, ..., \frac{1}{n}, -\frac{1}{2^n},...$
Where the positive terms form the harmonic series, and the negative terms form a convergent geometric series. Formally, the sequence is defined by
$a_n = \frac{1}{(n + 1)/2}$ if $n$ is odd
$a_n = -\frac{1}{2^{n / 2}}$ if $n$ is even
$\sum\limits_{n = 1}^\infty a_n$ will diverge to $\infty$.
A: A counterexample can be found on the Wikipedia Page for the Alternating Series Test.  Consider the series $$\sum_{n=4}^{\infty}(-1)^{n}\frac{1}{\sqrt{\lfloor \frac{n}{2}\rfloor}-1} = \frac{1}{\sqrt{2}-1} - \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} - 1} - \frac{1}{\sqrt{3}+1} + \dotsb.\tag{$\star$}$$  Then we have that $$\lim_{n\to\infty}\frac{1}{\sqrt{\lfloor \frac{n}{2}\rfloor}-1} = 0,$$ but it does not decrease monotonically, and it can be shown that the sum does not converge.  Indeed, note that in general $$\frac{1}{\sqrt{n}-1}-\frac{1}{\sqrt{n}+1} = \frac{\sqrt{n}+1-\sqrt{n}+1}{n-1} = \frac{2}{n-1}$$ and so the sum of the first $2m$ terms is given by $$S_{2m} = \frac{2}{1} + \frac{2}{2} + \frac{2}{3} + \dotsb + \frac{2}{m-1} >\sum_{i=1}^{m-1}\frac{1}{i}.$$  So, the partial sums for $(\star)$ exceed the partial sums for the harmonic series, which we know to diverge, meaning that $(\star)$ must diverge as well.
