determine if the series converge or diverge How can I go about determining if the following series converge or diverge.
$$\sum_{n=1}^{\infty} (-1)^n a_n$$

where
$$a_n=\begin{cases}
1/n,  &amp; \text{if $n$ is odd} \\
1/n^2, &amp; \text{if $n$ is even}
\end{cases}$$
Since
$$S=\sum_{n=1}^{\infty}(-1)^n a_n=\sum_{k=1}^{\infty}-\frac{1}{2k-1}+\sum_{k=1}^{\infty}\frac{1}{(2k)^2}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k-1/2}$$
And noting that the first term of the right side is less than 1
$$-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k-1/2}&lt;S&lt;-\frac{1}{2}\sum_{k=2}^{\infty}\frac{1}{k-1/2}$$
And since both of those series diverge, then $S$ is also divergent.
Since both the odd an even subsequences of the sequence $(a_{n})$ converge to zero, we can conclude that the entire sequence converges to zero.

There is a well-known result on real analysis which says that such series converge.

Lemma: If the sequence $(a_{n})_{n=1}^{\infty}$ converges to $0$, then the series
$$
\sum_{n=1}^{\infty}(-1)^{n}a_{n}
$$
converges.

You can find the proof of such fact on any book of Real Analysis. Actually, you can find it here.