Bounds or a nice form of the the partial sums of $\sum_{n=1}^\infty \frac{1}{n^2}$ I do know that $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, but so far I have not been able to found is either nice closed form for the sums $\sum_{n=K}^M \frac{1}{n^2}$ where $M > K \geq 1$, or alternatively some tight bounds, maybe in terms of $M$. By nice I mean that sure, you could expand each of the terms by hand, but then you would end up with a quite messy formula having a sum of products in the numerator and a quite clean product in the denominator.
So I am asking for a reference/sketch for a nice exact sum (if such even exists) or alternatively some tight upper and lower bound, preferably with some explanation. Thanks!
Edit: I feel dumb for not including this initially, but I was reading an algebra proof which used the bound $\sum_{i=N+1}^{M}\frac{1}{i^2} < \frac{1}{N^2}$ I got interested in knowing more about the bounds of the series in general. This means, unfortunately, that my specific question is a bit loose. Therefore my title could be: "What cool bounds/asymptotic connections do you know for the partial sums of the series $\sum_{n=1}^\infty \frac{1}{n^2}$?
 A: Using partial summation, it's not hard to see that
$$\sum_{n=1}^M \frac1{n^2}=\frac{\pi^2}6-\frac1M+\frac1{2M^2}+O\bigg(\frac1{M^3}\bigg).$$
Is this the sort of thing you were after?

If you want the lower-limit of the sum to be $n=K$, you can do
\begin{align*}
\sum_{n=K}^M\frac1{n^2}&=\sum_{n=1}^M\frac1{n^2}-\sum_{n=1}^{K-1}\frac1{n^2}\\[5pt]
&=\frac1K-\frac1M+\frac1{2K^2}+\frac1{2M^2}+O\bigg(\frac1{K^3}\bigg).
\end{align*}
A: For a very elementary bound, observe that $$\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}.$$
Therefore, by telescoping sum,
$$\sum_{n=K}^M \frac{1}{n^2} < \sum_{n=K}^M \left(\frac{1}{n-1} - \frac{1}{n} \right) = \frac{1}{K-1} - \frac{1}{M}.$$
Similarly,
$$\sum_{n=K}^M \frac{1}{n^2} > \sum_{n=K}^M \left(\frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{K} - \frac{1}{M+1}.$$
A: Not sure if this is what you want, but it is too long for a comment, so here we go: You can write it in terms of the derivative of the digamma function, i.e. the second derivative of $\log \circ \Gamma$. For this note that
$$f(z) := (\log \circ \Gamma)''(z) = \frac{\mathrm{d}}{\mathrm{d}z}\frac{\Gamma'(z)}{\Gamma(z)} = \frac{\Gamma''(z)\Gamma(z) - \Gamma'(z)^2}{\Gamma(z)^2}$$
so that
\begin{align*}
f(z+1) &= \frac{\Gamma''(z+1)\Gamma(z+1) - \Gamma'(z+1)^2}{\Gamma(z+1)^2}\\
&= \frac{(2\Gamma'(z) + z\Gamma''(z))z\Gamma(z) - (z\Gamma'(z) + \Gamma(z))^2}{z^2\Gamma(z)^2}\\
&= \frac{z^2\Gamma(z)\Gamma''(z) - z^2\Gamma'(z)^2 - \Gamma(z)^2}{z^2\Gamma(z)^2}\\
&= f(z) - \frac{1}{z^2}
\end{align*}
Thus, since $f(1) = \frac{\pi^2}{6}$, we have
$$\sum_{n = 1}^M \frac{1}{n^2} = \frac{\pi^2}{6} - f(M+1).$$
That is, $f(M+1)$ is actually equal to the tail of the series. For $\sum\limits_{n = K}^M \frac{1}{n^2}$ you can then get
$$\sum\limits_{n = K}^M \frac{1}{n^2} = f(K) - f(M+1),$$
of course.
