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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.

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    $\begingroup$ Do you have to use this series? There are other series which converge more quickly... $\endgroup$ – abiessu Sep 12 '17 at 0:50
  • $\begingroup$ I know. But I just want to crack it. $\endgroup$ – poisson Sep 12 '17 at 0:54
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    $\begingroup$ $$ζ(2) = \frac{\pi^2}{6} = \sum_{n = 1}^{\infty}\frac{3}{n^2{{2n}\choose{n}}}$$ $\endgroup$ – Feeds Sep 12 '17 at 1:26
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    $\begingroup$ Have a look at the first sections of these notes (goo.gl/Atr6K7) where the identity provided by user477343 is proved through many different approaches (creative telescoping, logarithmic integrals, Fourier-Legendre series, complex Analysis...) $\endgroup$ – Jack D'Aurizio Sep 12 '17 at 17:06
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    $\begingroup$ Ah, well I hope your time becomes $C^\infty$ @JackD'Aurizio $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 20:28
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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,

$$\frac{\pi^2}6=\sum_{k=0}^\infty\frac1{2^k}\sum_{n=0}^k\binom kn\frac{(-1)^n}{(n+1)^2}$$

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    $\begingroup$ I guess that $\binom{n}{k}$ should be $\binom{k}{n}$. It is also worth noticing that the resulting series from Euler series acceleration converges exponentially fast, due to the bound $$0\leq \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{(k+1)^s} \leq 1$$ which holds for all $s>0$ and $n\geq 0$. $\endgroup$ – Sangchul Lee Sep 12 '17 at 1:48
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    $\begingroup$ Oops, yes, creature of habit, I like to have $\binom nk$. $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 1:50
  • $\begingroup$ O.O I had never realized how efficient EM formula is here $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 1:54

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