How to accelerate the convergence of $1 + \frac{1}{2^2} + \frac{1}{3^2} + \ldots$? It is well known that 
$$ \frac{\pi^2}{6} = 1 + \frac{1}{4} +\frac{1}{9} + \frac{1}{16} + \ldots  $$
I am trying to use it to calculate $\pi $. The problem is how to accelerate the convergence of the series on the right hand side. 
The Shanks transform does not work because it converges only in a power law, not exponentially. 
 A: By the Euler-Maclaurin formula,
$$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^{a-1}\frac1{n^2}+\underbrace{\int_a^\infty\frac1{x^2}~\mathrm dx}_{=1/a}+\frac1{2a^2}+\sum_{k=1}^p\frac{B_{2k}}{a^{2k+1}}+R_p$$
$$|R_p|\le\frac{4(2p)!}{(6a)^{2p+1}}$$
Choosing large enough $a$ will result in very fast convergence (Euler himself apparently evaluated this series to 20 places, according to the Wikipedia link above)
We may easily do the same, by taking $a=10$ and $p=5$,
$$\frac{\pi^2}6=\tiny1+\frac14+\frac19+\frac1{16}+\frac1{25}+\frac1{36}+\frac1{49}+\frac1{64}+\frac1{81}+\frac1{10}+\frac1{200}+\frac1{6000}-\frac13\times10^{-6}+\frac1{42}\times10^{-7}-\frac13\times10^{-10}+\frac1{132}\times10^{-10}\pm10^{-23}$$

By alternating it:
$$S_+=\sum_{n=1}^\infty\frac1{n^2}$$
$$S_-=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}$$
Subtract them and you get
$$S_+-S_-=\sum_{n=1}^\infty\frac{1+(-1)^n}{n^2}=\sum_{n=1}^\infty\frac2{(2n)^2}=\frac12S_+$$
Thus,
$$S_+=2S_-$$
$S_-$ may then be accelerated using an Euler transform,
$$S_-=\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kn\frac{(-1)^n}{(n+1)^2}$$
That is,
$$\frac{\pi^2}6=\sum_{k=0}^\infty\frac1{2^k}\sum_{n=0}^k\binom kn\frac{(-1)^n}{(n+1)^2}$$
