<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] Thomas Andrews is right when he says that there are no general rules. Often this type of problem, we must use a more intuitive method.
I know the p-series for p = 2. The resolution method in this series comes from Euler himself.
(1) $\sin(x) = x - x^3/3! + x^5/5! -x^7/7! + ...$ (Taylor series)
(2) $\sin(x)/x = 1 - x^2/3! + x^4/5! -x^6/7! + ...$ (divided by x)
(3) Now, the roots (intersections with the x-axis) of sin(x)/x occur precisely at x = $n\pi$ where n = ±1,±2,±3,...  Let us assume we can express this infinite series as a (normalized) product of linear factors given by its roots, just as we do for finite polynomials
\begin{align} \frac{\sin(x)}{x} & {} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ & {} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{align} 
(4) If we formally multiply out this product and collect all the $x^2$ terms (we are allowed to do so because of Newton's identities), we see that the $x^2$ coefficient of sin(x)/x is
$$
1/\pi^2 + 1/(4\pi^2) + 1/(9\pi^2) + ... = (1/\pi^2)\sum_{n=1}^\infty\frac{1}{n^2}
$$
(5) But from the original infinite series expansion of sin(x)/x, the coefficient of $x^2$ is −1/(3!) = −1/6. These two coefficients must be equal...
$$
-1/6 = -(1/\pi^2)\sum_{n=1}^\infty\frac{1}{n^2}
$$
(6) Multiplying through both sides of this equation by $-\pi^2$ gives the sum of the reciprocals of the positive square integers.
$$
\sum_{n=1}^\infty\frac{1}{n^2} = \pi^2/6
$$
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