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

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?

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

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Very neat solution! +1

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Thank you very much ! :-)

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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 \(\frac{x^2}{\sqrt{1-x^2}}\)?

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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

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

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You beated me while writing the solution.

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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....

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

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

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