Compute $\sum \frac{1}{k^2}$ using Euler-Maclaurin formula I read that Euler used the summation formula to calculate the value of the series $\sum_{k =1}^{\infty} \frac{1}{k^2}$ to high precision without too much hassle. The article Dances between continuous and discrete: Euler’s summation formula goes into the calculation, however without too much justification of why it works (especially since the series used to calculate the limit does not converge and one has to truncate it at a certain point). I would be glad if someone could elaborate from a more modern viewpoint on how and why it works.
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Numerically, we must sum up a 'few' terms before we use Euler-Maclaurin ( in the second term ).
For example: 
\begin{align}
\sum_{n = 1}^{\infty}{1 \over n^{2}}
&=\sum_{n = 1}^{N}{1 \over n^{2}} + \sum_{n = 1}^{\infty}{1 \over \pars{n + N}^{2}}
\approx\sum_{n = 1}^{N}{1 \over n^{2}}
+\int_{0}^{\infty}{\dd x \over \pars{x + N}^{2}}
=\sum_{n = 1}^{N}{1 \over n^{2}}
+ {1 \over N}
\end{align}

This '$\ds{\tt\mbox{extremely simple formula}}$', for example, yields a relative error of
  $\ds{1.14\ \%}$ with $\ds{N = 5}$.

Historically; it seems Euler was convinced, by means of the Euler-Maclaurin formula, that the correct value was $\ds{\pi^{2} \over 6}$ before he tried to demonstrate it.
