π\pi!(2)

π2=4+32n=1n(2n1)(2n+1)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  x21x2\ \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 Hamza A
3 years, 6 months ago

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1 vote

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Using partial fractions, we can observe that: 32n(2n1)(2n+1)2=42n142n+1+8(2n+1)2\frac{32n}{(2n-1)(2n+1)^2} = \frac{4}{2n-1} - \frac{4}{2n+1} + \frac{8}{(2n+1)^2}

Now n=1(42n142n+1) \sum_{n=1}^{\infty} \left(\frac{4}{2n-1} - \frac{4}{2n+1}\right) is a telescoping sum, whose value is 44. So there is just one more sum to calculate: n=18(2n+1)2=8(k=11k2n=11(2n)2112)\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)=8π268π2468=π28= \frac{8\pi^2}{6} -\frac{8\pi^2}{4*6} - 8 = \pi^2 - 8

Therefore, 4+32n=1n(2n1)(2n+1)2=4+(4+π28)=π24+32\sum_{n=1}^{\infty} \frac{n}{(2n-1)(2n+1)^2} = 4+(4+\pi^2-8) = \pi^2QED

I'm curious how you got this from the series of x21x2\frac{x^2}{\sqrt{1-x^2}}?

Ariel Gershon - 3 years, 6 months ago

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so ,i used the expansion of x21x2\frac { { x }^{ 2 } }{ \sqrt { 1-{ x }^{ 2 } } } which is n=1(2n2n2)4n1x2n\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=sintx=\sin{t} and i got

n=1(2n2n2)4n1sin2n+1t2n+1=tsintcost2\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 00 to π2\frac{\pi}{2} and got with a lot of rearranging and simplifying and comparing with other sums i get π2=4+32n=1n(2n1)(2n+1)2 { \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

π21614=12n=1Γ(n+1)π(2n2n1)4n1(2n+1)Γ(n+32)\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

Hamza A - 3 years, 6 months ago

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

Ariel Gershon - 3 years, 6 months ago

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@Ariel Gershon your method was way simpler,i should've used that one ;)

Hamza A - 3 years, 6 months ago

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You beated me while writing the solution.

Akshat Sharda - 3 years, 6 months ago

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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....

Anshuman Bais - 3 years, 6 months ago

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@Anshuman Bais Yeah !! You are right !!!! :P

Akshat Sharda - 3 years, 6 months ago

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

Ariel Gershon - 3 years, 6 months ago

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

Hamza A - 3 years, 6 months ago

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4+32n=1n(2n1)(2n+1)24+32n=1(18(2n1)18(2n+1)+14(2n+1)2)4+n=1(4(2n1)4(2n+1)+8(2n+1)2)4+(4+43+45+4345+8n=11(2n+1)2n=11n2n=11(2n)21=π26π2241=π281)8+8(π281)8+π28=π2 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 - 3 years, 6 months ago

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Very neat solution! +1

Pi Han Goh - 3 years, 6 months ago

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Thank you very much ! :-)

Akshat Sharda - 3 years, 6 months ago

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