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!
 A: 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, $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2} = \displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} - \displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{8} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{12}$. 
A: If we did not know an exact answer, here is a crude way to estimate the sum of the series.  
Our series is an alternating series, with terms that decrease in absolute value and have limit $0$. So the error made by truncating at a particular term has absolute value less than the absolute value of the first "neglected" term. 
For example, if we use $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}$ as an approximation, then the absolute value of the error is less than $\frac{1}{6^2}$. Furthermore, the error is "negative," that is, our estimate is greater than the true value.  
A: 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.
A: HINT:
This can be written as $$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i^2}$$
$$$$
Setting $\displaystyle a_i= \frac{(-1)^{i+1}}{i^2}$:
$$\sum_{i=1}^{\infty} a_{2i-1}=\sum_{i=1}^{\infty} \frac{1}{(2i-1)^2}$$
$$\sum_{i=1}^{\infty} a_{2i}=-\frac{1}{4}\sum_{i=1}^{\infty} \frac{1}{i^2}$$
When the series $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}$ converge then the series $\displaystyle \sum_{i=1}^{\infty} a_{i}$ also converges.
$$\sum_{i=1}^{\infty} a_{i}=\sum_{i=1}^{\infty} a_{2i-1}+\sum_{i=1}^{\infty} a_{2i}$$
Find where the series $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}$ converge.
