I read this in a book somewhere, but I was looking for it just now and I couldn't find the book. I also couldn't find any sort of reference anywhere else. So I'm wondering if this is well-known, and if so, where I can find a proof of it.
Is it true that $\sum_{i=1}^k\frac{1}{i^2}=\frac{\pi^2}{6}-\sum_{j=0}^{\infty}\frac{B_j}{k^{j+1}}$ where $B_j$ is the $j$'th Bernoulli number, with the convention $B_1=-\frac{1}{2}$? That is, is $1+\frac{1}{4}+\ldots+\frac{1}{10000^2}=\frac{\pi^2}{6}-\frac{1}{10000}+\frac{1}{2\cdot10000^2}-\frac{1}{6\cdot10000^3}+\ldots$?
It appears to be the case on Mathematica, checking this exact case with 50 digits of accuracy. But I've never seen this simple equation anywhere before. I also don't know enough about the Bernoulli numbers to do anything except spot them when they appear.