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

Very neat solution! +1 · 9 months, 1 week ago

Thank you very much ! :-) · 9 months, 1 week ago

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 $$\frac{x^2}{\sqrt{1-x^2}}$$? · 9 months, 1 week ago

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 · 9 months, 1 week ago

Wow, looks like a lot of work! Nice job though! · 9 months, 1 week ago

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

You beated me while writing the solution. · 9 months, 1 week 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.... · 9 months, 1 week ago

Yeah !! You are right !!!! :P · 9 months, 1 week ago

Interesting, we both had the same solution at the same time · 9 months, 1 week ago