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 .

#Calculus #RiemannZetaFunction #Convergence #SummationOfSeries

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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

حكيم الفيلسوف الضائع - 6 years, 11 months ago

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 - 6 years, 10 months ago

Alright thanks!

حكيم الفيلسوف الضائع - 6 years, 10 months ago

@Pratik Shastri @Ronak Agarwal @Kishlaya Jaiswal @Azhaghu Roopesh M

U Z - 6 years, 3 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.

Haroun Meghaichi - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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

Comments

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