Show series $1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2$ ... does not converge

Question

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.

Answer 1

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.

Answer 2

First, there seems to be quite a bit of confusion in the comments concerning "grouping". Let's take a look at it, essentially:

\begin{align} 1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} +\cdots &\stackrel{?}{=} \left(1 - \frac{1}{2^2}\right) + \left(\frac{1}{3} - \frac{1}{4^2}\right) + \left(\frac{1}{5} - \frac{1}{6^2}\right) + \cdots \\ &= \sum_{n=0}^\infty \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right). \end{align}

Now, when you say that you're "not allowed to group", I guess you're right in that you've generated a somewhat different series but the two series are concretely related. In particular, if the left side converges, then the right side must also converge. This is a special case of the fact that, if a sequence converges, then any subsequence of that sequence also converges. Specifically, if we let $s_n$ denote the $n^{\text{th}}$ partial sum of the series on the left and we let

$$S_n = \sum_{n=0}^n \left(\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2}\right)$$

denote the $n^{\text{th}}$ partial sum of the series on the right, then $S_n = s_{2n}$. Thus, $S_n$ is a subsequence of $s_n$ and, if $s_n$ converges then $S_n$ must converge to the same limit. Taking the contrapositive, if $S_n$ diverges, then $s_n$ must also diverge.

I think you make a mistake, though, at the next step by breaking the series up into two series. You are essentially rearranging the series which is only valid when the series is absolutely convergent.

The approach at this point is to simply combine the fractions to get $$\frac{1}{2n + 1} - \frac{1}{(2n + 2)^2} = \frac{4 n^2+6 n+3}{4 (n+1)^2 (2 n+1)},$$ to which the limit comparison test is easily applicable.

This is exactly the approach I took in my answer to this question.