Speeding up the calculation of ζ(2)\zeta(2).

A year ago, I have posted this article, it states that if you want the correct result close by 0.010.01 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 ζ(2)=2n>0(1)n+1n2\zeta(2)=2\sum_{n>0} \frac{(-1)^{n+1}}{n^2}, our goal is to see how much is it faster. We set : am=n=1m1n2     ,     bm=2n=1m(1)n+1n2     ,m>1a_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 : nζ(2)bnζ(2)an1n \left|\frac{\zeta(2) -b_n}{\zeta(2) - a_n} \right|\to 1 This means that (bn)(b_n) converge a lot quicker than (an)(a_n).

Image credit : Recovering Lutheran blog .

Note by Haroun Meghaichi
6 years, 6 months ago

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

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Yes of course, check this page.

Euler did not know all these fancy series (but the first) but he managed to get 2020 decimals using the Euler-Maclaurin summation formula which is also better than bnb_n.

Haroun Meghaichi - 6 years, 6 months ago

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Alright thanks!

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@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.

Haroun Meghaichi - 6 years, 5 months ago

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