What's the sum of this series? I would like to know how to find out the sum of this series:
$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$
The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum? 
Thanks!

It is well known that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$. 
Thus, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \sum_{m=1}^{\infty}\dfrac{1}{(2m)^2} = \dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2} = \dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{24}$. 
Hence, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} = \sum_{n=1}^{\infty}\dfrac{1}{n^2} - \sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{8}$. 
Finally, $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2} = \displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} - \displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{8} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{12}$. 
