# Speeding up the calculation of $\zeta(2)$.

A year ago, I have posted this article, it states that if you want the correct result close by $0.01$ you should calculate about 100 terms which is huge by hand (Euler hadn't W|A to help him at the time he Calculated it ).

A way to speed up the convergence is to use $\zeta(2)=2\sum_{n>0} \frac{(-1)^{n+1}}{n^2}$, our goal is to see how much is it faster. We set : $a_m= \sum_{n=1}^{m} \frac{1}{n^2} \ \ \ \ \ , \ \ \ \ \ b_m=2\sum_{n=1}^{m} \frac{(-1)^{n+1}}{n^2}\ \ \ \ \ , m>1$

Prove or disprove that : $n \left|\frac{\zeta(2) -b_n}{\zeta(2) - a_n} \right|\to 1$ This means that $(b_n)$ converge a lot quicker than $(a_n)$.

Image credit : Recovering Lutheran blog .

Note by Haroun Meghaichi
7 years ago

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Are there any established series that converge faster than $(bn)$?

Euler did not know all these fancy series (but the first) but he managed to get $20$ decimals using the Euler-Maclaurin summation formula which is also better than $b_n$.

- 7 years ago

Alright thanks!

- 6 years, 5 months ago

@Pranav Arora ,@Michael Lee ,@Tunk-Fey Ariawan, @Michael Mendrin (and all calculus guys). What do you think of this problem ? I can post hints if you want.

Another thing, what kind of calculus problems do prefer ? AFAIS, it is integration.

- 7 years ago