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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
3 years, 5 months ago

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

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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 \(20\) decimals using the Euler-Maclaurin summation formula which is also better than \(b_n\).

Haroun Meghaichi - 3 years, 5 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 - 3 years, 5 months ago

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