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 .

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

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 \(b_n\).

Log in to reply

Alright thanks!

Log in to reply

@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.

Log in to reply