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

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.

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TopNewest\( 4+32\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n }{ (2n-1) (2n+1)^{ 2 } } } \\ 4+32 \displaystyle \sum^{\infty}_{n=1}\left( \frac{1}{8(2n-1)}-\frac{1}{8(2n+1)}+\frac{1}{4(2n+1)^2}\right) \\ 4+ \displaystyle \sum^{\infty}_{n=1} \left( \frac{4}{(2n-1)}-\frac{4}{(2n+1)}+\frac{8}{(2n+1)^2}\right) \\ 4+\left( 4+\frac{4}{3}+\frac{4}{5}+\ldots -\frac{4}{3}-\frac{4}{5}-\ldots +8\underbrace{\displaystyle \sum^{\infty}_{n=1} \frac{1}{(2n+1)^2}}_{\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2}-\displaystyle \sum^{\infty}_{n=1}\frac{1}{(2n)^2}-1 = \frac{\pi^2}{6}-\frac{\pi^2}{24}-1=\frac{\pi^2}{8}-1 } \right) \\ 8+8\left(\frac{\pi^2}{8}-1\right) \\ 8+\pi^2-8=\boxed{\pi^2}\) – Akshat Sharda · 11 months ago

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– Pi Han Goh · 10 months, 4 weeks ago

Very neat solution! +1Log in to reply

– Akshat Sharda · 10 months, 4 weeks ago

Thank you very much ! :-)Log in to reply

Using 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 \(\frac{x^2}{\sqrt{1-x^2}}\)? – Ariel Gershon · 11 months ago

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

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 – Hummus A · 11 months ago

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– Ariel Gershon · 11 months ago

Wow, looks like a lot of work! Nice job though!Log in to reply

– Hummus A · 11 months ago

your method was way simpler,i should've used that one ;)Log in to reply

– Akshat Sharda · 11 months ago

You beated me while writing the solution.Log in to reply

– Anshuman Bais · 11 months ago

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

– Akshat Sharda · 11 months ago

Yeah !! You are right !!!! :PLog in to reply

– Ariel Gershon · 11 months ago

Interesting, we both had the same solution at the same timeLog in to reply

– Hummus A · 11 months ago

i'll post how once i get back from school :)Log in to reply