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!

The sum of this series is $\frac{\pi^2}{12}$.
Explantion
We already know that:
$$1+ \frac{1}{2^2} + \frac{1}{3^2} + \text{...} = \frac{\pi^2}{6}$$
HINT
Note that:
$$\large \frac{-1}{2^2} = \frac{1}{2^2} - \frac{1}{2^2} - \frac{1}{2^2}$$
Now,
$$1- \frac{1}{2^2} + \frac{1}{3^2} - \text{...} = \left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right) - 2 * \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + ...\right)$$ 
$$ = \frac{\pi^2}{6} - \frac{2}{2^2}\left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right)$$
$$ = \frac{\pi^2}{6} - \frac{1}{2}*\frac{\pi^2}{6}$$
$$= \frac{\pi^2}{12}$$
Comment if you have questions.
