<s>[INST] Show series $1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2$ ... does not converge I was wondering if my proof is correct, and if there are any better alternative proofs. Or maybe proof that use nice tricks i might need in the future.
$$1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} \ldots = \sum_{n =1}^\infty \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right) + 1 - \frac{1}{2^2}$$
Now we know that $$\sum_{n = 1}^\infty \frac{1}{2n + 1}$$ diverges and $$\sum_{n = 1}^\infty -\frac{1}{(2n + 2)^2}$$ converges. Hence their sum diverges (I proved this fact). Hence, the series diverges. Any obvious mistake or better way of tackling it? Maybe using partial sums since i am clueless how to use them. 
 [/INST] Call $s_k$ the partial sums of the series. That is $s_k=\sum_{n=1}^ka_n$, where $a_n=\frac{1}{n}$ if $n$ is odd, and $a_n=-\frac{1}{n^2}$ if $n$ is even. As $\sum_{j=1}^\infty \frac{1}{j^2}=\frac{\pi^2}{6}$, we have the estimate
$$
s_k\geq-\frac{\pi^2}{6}+\sum_{n=1}^ka_{2n-1}=-\frac{\pi^2}{6}+\sum_{n=1}^k\frac{1}{2n-1}.
$$
By comparison with the harmonic series the last sum is divergent (when $k\to\infty$) so $s_k$ must be too, proving that your series diverges.
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