<s>[INST] Which (convergent) series can one find the sum of? I know about geometric series and how one can find the sum when they are convergent.
I also have heard that one can prove that the $p$-series 
$$\sum_{n=1}^{\infty} \frac{1}{n^2}$$
has sum $\pi^2/6$ (but I don't know how this is actually proved).
This might be an unanswerable question, but are there any tools or general rules that will tell you when you can find the exact sum of a given series? Are there, for example, rules/tools for how to find the exact value of any $p$-series? (like with geometric series).
 [/INST] This is a famous sum & there are multiple ways of approaching this. Here is one way via a geometric sum:
We have the well known sum
\begin{equation*}
1+x+x^2+\cdots =\frac{1}{1-x}.
\end{equation*}
Setting $x=e^{i\theta},~0<\theta<2\pi~(x\neq 1)$ gives
\begin{equation*}
1+e^{i\theta}+e^{2i\theta}+\cdots =(1-e^{i\theta})^{-1}=\frac{1}{2}+\frac{1}{2}i\cot(\frac{1}{2}\theta).
\end{equation*}
Equating the real parts we see that
\begin{equation*}
\frac{1}{2}+\cos(\theta)+\cos(2\theta)+\cdots =0.
\end{equation*}
Replacing $\theta$ with $\theta+\pi$ & using the trigonometric addition formulae we get
\begin{equation*}
\frac{1}{2}-\cos(\theta)+\cos(2\theta)-\cdots =0.
\end{equation*}
Integrate from $\theta=0$ to $\theta=\phi$ & write $\theta$ again for $\phi$ we obtain
\begin{equation*}
\sin(\theta)-\frac{1}{2}\sin(2\theta)+\frac{1}{3}\sin(3\theta)-\cdots =\frac{1}{2}\theta,~-\pi<\theta<\pi
\end{equation*}
& this series is convergent. Integrating again gives
\begin{equation*}
1-\cos(\theta)-\frac{1-\cos(2\theta)}{2^2}+\frac{1-\cos(3\theta)}{3^2}-\cdots =\frac{1}{4}\theta^2.
\end{equation*}
Setting $\theta=\pi$ gives 
\begin{equation*}
1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots =\frac{1}{8}\pi^2.
\end{equation*}
Since 
\begin{equation*}
1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots =1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots -\frac{1}{2^2}-\frac{1}{4^2}-\cdots
=(1-\frac{1}{4})(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots),
\end{equation*}
we deduce that 
\begin{equation*}
1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots = \sum^{\infty}_{n=1} \frac{1}{n^2} =\frac{1}{6}\pi^2.
\end{equation*}
Remark: We also get the formula
\begin{equation*}
1-\frac{1}{2^2}+\frac{1}{3^2}-\cdots =\frac{1}{12}\pi^2
\end{equation*}
</s>