Wolfram Alpha result for infinite series summation .
Sebastian Wright 
Consider the infinite sum $s=1+1/2^2-1/3^2-1/4^2+1/5^2+1/6^2-...$. We can see that the series is absolutely convergent and hence convergent. But WolframAlpha seems to give me a different answer. When I add two more terms it shows me that the series converges. Can anyone explain why this happens and also give an analytical expression for the sum ?
2 Answers
$\begingroup$Wolfram alpha guessed your general term wrong. If you look at the cell titled "Result", you see the general term $$\frac{-54-(-7+n)n(16+(-7+n)n)}{n^2}$$ instead of what you intended.
Try instead inputting the sum directly.
We can compute the sum explicitly by computing four related sums using Wolfram alpha: $$ \begin{align*} S_0 &= \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}, \\ S_1 &= \sum_{n=1}^\infty \frac{i^n}{n^2} = -\frac{\pi^2}{48} + Ci, \\ S_2 &= \sum_{n=1}^\infty \frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}, \\ S_3 &= \sum_{n=1}^\infty \frac{(-i)^n}{n^2} = -\frac{\pi^2}{48} - Ci, \end{align*} $$ where $C$ is Catalan's constant. The sum we are after is $$ \left(-\frac{1}{2}-\frac{1}{2}i\right)S_1 + \left(-\frac{1}{2}+\frac{1}{2}i\right)S_3 = \frac{\pi^2}{48} + C. $$
$\endgroup$ 3 $\begingroup$$$\sum_{n\geq0}\left(-1\right)^{n}\left(\frac{1}{\left(2n+2\right)^{2}}+\frac{1}{\left(2n+1\right)^{2}}\right)=\sum_{n\geq0}\frac{\left(-1\right)^{n}}{\left(2n+2\right)^{2}}+\sum_{n\geq0}\frac{\left(-1\right)^{n}}{\left(2n+1\right)^{2}}=\frac{1}{4}\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n^{2}}+C=\frac{1}{4}\eta\left(2\right)+C$$ where $C$ is the Catalan constant and $\eta$ is the Dirichlet eta funcion. Using the identity$$\eta\left(s\right)=\left(1-2^{1-s}\right)\zeta\left(s\right)$$ with $$\textrm{Re}\left(s\right)>1,$$ and $\zeta$ is the Riemann zeta function we have$$\eta\left(2\right)=\frac{1}{2}\zeta\left(2\right)=\frac{\pi^{2}}{12}$$ so$$\sum_{n\geq0}\left(-1\right)^{n}\left(\frac{1}{\left(2n+2\right)^{2}}+\frac{1}{\left(2n+1\right)^{2}}\right)=\frac{\pi^{2}}{48}+C.$$
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