\[ { \pi }^{ 2 }=4+32\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n }{ (2n-1)(2n+1)^{ 2 } } } \]

I was playing around with the expansion of \(\ \dfrac { { x }^{ 2 } }{ \sqrt { 1-{ x }^{ 2 } } } \) this time, and I found the series above. Can you prove it?

For a related question, see this.

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:

TopNewestUsing partial fractions, we can observe that: \[\frac{32n}{(2n-1)(2n+1)^2} = \frac{4}{2n-1} - \frac{4}{2n+1} + \frac{8}{(2n+1)^2}\]

Now \( \sum_{n=1}^{\infty} \left(\frac{4}{2n-1} - \frac{4}{2n+1}\right)\) is a telescoping sum, whose value is \(4\). So there is just one more sum to calculate: \[\sum_{n=1}^{\infty} \frac{8}{(2n+1)^2} = 8\left(\sum_{k=1}^{\infty} \frac{1}{k^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2} - \frac{1}{1^2}\right)\]\[= \frac{8\pi^2}{6} -\frac{8\pi^2}{4*6} - 8 = \pi^2 - 8\]

Therefore, \[4+32\sum_{n=1}^{\infty} \frac{n}{(2n-1)(2n+1)^2} = 4+(4+\pi^2-8) = \pi^2\]

QEDI'm curious how you got this from the series of \(\frac{x^2}{\sqrt{1-x^2}}\)?

Log in to reply

so ,i used the expansion of \(\frac { { x }^{ 2 } }{ \sqrt { 1-{ x }^{ 2 } } } \) which is \(\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { \begin{pmatrix} 2n-2 \\ n-2 \end{pmatrix} }{ { 4 }^{ n-1 } } } { x }^{ 2n }\) integrated both sides then put \(x=\sin{t}\) and i got

\(\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { \begin{pmatrix} 2n-2 \\ n-2 \end{pmatrix} }{ { 4 }^{ n-1 } } } \frac { { \sin^{2n+1}{t} } }{ 2n+1 } =\frac { t-\sin { t } \cos { t } }{ 2 } \)

then i integrated from \(0\) to \(\frac{\pi}{2}\) and got with a lot of rearranging and simplifying and comparing with other sums i get \[ { \pi }^{ 2 }=4+32\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n }{ (2n-1)(2n+1)^{ 2 } } } \]

this is how it looked before all the comparing and substituting

\(\frac { { \pi }^{ 2 } }{ 16 } -\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { \Gamma (n+1)\sqrt { \pi } \begin{pmatrix} 2n-2 \\ n-1 \end{pmatrix} }{ { 4 }^{ n-1 }(2n+1)\Gamma (n+\frac { 3 }{ 2 } ) } } \)

this was a practice problem in my textbook,so i had faith in solving it ;),it was a bit tedious though

Log in to reply

Wow, looks like a lot of work! Nice job though!

Log in to reply

Log in to reply

You beated me while writing the solution.

Log in to reply

Two different Countries, two different Timezones but then also you two, writing the same solution of the same question at the same time. Wow! These is possible on Brilliant only. Cheers....

Log in to reply

Log in to reply

Interesting, we both had the same solution at the same time

Log in to reply

i'll post how once i get back from school :)

Log in to reply

\( 4+32\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n }{ (2n-1) (2n+1)^{ 2 } } } \\ 4+32 \displaystyle \sum^{\infty}_{n=1}\left( \frac{1}{8(2n-1)}-\frac{1}{8(2n+1)}+\frac{1}{4(2n+1)^2}\right) \\ 4+ \displaystyle \sum^{\infty}_{n=1} \left( \frac{4}{(2n-1)}-\frac{4}{(2n+1)}+\frac{8}{(2n+1)^2}\right) \\ 4+\left( 4+\frac{4}{3}+\frac{4}{5}+\ldots -\frac{4}{3}-\frac{4}{5}-\ldots +8\underbrace{\displaystyle \sum^{\infty}_{n=1} \frac{1}{(2n+1)^2}}_{\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2}-\displaystyle \sum^{\infty}_{n=1}\frac{1}{(2n)^2}-1 = \frac{\pi^2}{6}-\frac{\pi^2}{24}-1=\frac{\pi^2}{8}-1 } \right) \\ 8+8\left(\frac{\pi^2}{8}-1\right) \\ 8+\pi^2-8=\boxed{\pi^2}\)

Log in to reply

Very neat solution! +1

Log in to reply

Thank you very much ! :-)

Log in to reply