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\(\pi\)!(2)

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

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.

Note by Hummus A
9 months, 1 week ago

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\( 4+32\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n }{ (2n-1) (2n+1)^{ 2 } } } \\ 4+32 \displaystyle \sum^{\infty}_{n=1}\left( \frac{1}{8(2n-1)}-\frac{1}{8(2n+1)}+\frac{1}{4(2n+1)^2}\right) \\ 4+ \displaystyle \sum^{\infty}_{n=1} \left( \frac{4}{(2n-1)}-\frac{4}{(2n+1)}+\frac{8}{(2n+1)^2}\right) \\ 4+\left( 4+\frac{4}{3}+\frac{4}{5}+\ldots -\frac{4}{3}-\frac{4}{5}-\ldots +8\underbrace{\displaystyle \sum^{\infty}_{n=1} \frac{1}{(2n+1)^2}}_{\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2}-\displaystyle \sum^{\infty}_{n=1}\frac{1}{(2n)^2}-1 = \frac{\pi^2}{6}-\frac{\pi^2}{24}-1=\frac{\pi^2}{8}-1 } \right) \\ 8+8\left(\frac{\pi^2}{8}-1\right) \\ 8+\pi^2-8=\boxed{\pi^2}\) Akshat Sharda · 9 months, 1 week ago

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@Akshat Sharda Very neat solution! +1 Pi Han Goh · 9 months, 1 week ago

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@Pi Han Goh Thank you very much ! :-) Akshat Sharda · 9 months, 1 week ago

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

I'm curious how you got this from the series of \(\frac{x^2}{\sqrt{1-x^2}}\)? Ariel Gershon · 9 months, 1 week ago

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@Ariel Gershon 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 Hummus A · 9 months, 1 week ago

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@Hummus A Wow, looks like a lot of work! Nice job though! Ariel Gershon · 9 months, 1 week ago

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@Ariel Gershon your method was way simpler,i should've used that one ;) Hummus A · 9 months, 1 week ago

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@Ariel Gershon You beated me while writing the solution. Akshat Sharda · 9 months, 1 week ago

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@Akshat Sharda 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.... Anshuman Bais · 9 months, 1 week ago

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@Anshuman Bais Yeah !! You are right !!!! :P Akshat Sharda · 9 months, 1 week ago

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@Akshat Sharda Interesting, we both had the same solution at the same time Ariel Gershon · 9 months, 1 week ago

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@Ariel Gershon i'll post how once i get back from school :) Hummus A · 9 months, 1 week ago

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