TITLE: How to show the following infinite series is convergent
QUESTION [2 upvotes]: I am stuck on the following problem: 

I have to show that the following series 
  is convergent: $$1-\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}-\frac{1}{6^2}+\ldots$$ .

I was trying to make some rearrangement and trying to convert it in alternating series and apply "Leibnitz's test" but failed to do so.
Can someone explain? Thanks and regards to all.

REPLY [3 votes]: Thanks to T.Bongers and Mhenni Benghorbal. I think I have got the trick.
We see that
$$1-\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}-\frac{1}{5^2}-\frac{1}{6^2}+\cdots \le1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+ \cdots$$ $$= \sum\limits_{k = 1}^{\infty} \frac{1}{k^2}$$ and so It is absolutely convergent since $p = 2 > 1$;

REPLY [1 votes]: In leibnitz test you have to show that $u_n$ satiesfies 2 conditions:

$u_{n+1}-u_n<0$
$\lim_{x\rightarrow \infty} u_n=0$

Here $u_n=\frac{1}{n^2}$
It's really easy to show that the 2 tests are valid here.
$$\frac{1}{(n+1)^2}-\frac{1}{n^2}<0$$
and $$\lim_{x\rightarrow \infty} \frac{1}{x^2}=0$$
Hence the series $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^2}$ is convergent
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