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\(\pi\)!

\[ \large { \pi }^{ 2 }=\frac { 128 }{ 9 } -128\displaystyle\sum _{ n=1 }^{ \infty }{ \frac { n+1 }{ (2n+1)(2n-1)(2n+3)^{ 2 } } } \]

I was playing with the expansion of \(t^2\sqrt{1-t^2}\) and I came to the series above, can you prove it?

Note by Hummus A
9 months, 2 weeks ago

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Here is a quick method.

Using partial fractions, we have,

\(\displaystyle \sum_{n=0}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = \sum_{n=0}^{\infty} \left[-\dfrac{1}{16(2 n+1)} + \dfrac{1}{64(2 n+3)} - \dfrac{1}{16 (2 n+3)^2} + \dfrac{3}{64 (2 n-1)} \right] \)

By Regularization, we have,

\(\displaystyle = \dfrac{1}{32} \psi\left(\dfrac{1}{2}\right) - \dfrac{1}{128} \psi\left(\dfrac{3}{2}\right) -\dfrac{1}{64} \psi^{(1)} \left(\dfrac{3}{2}\right) + \dfrac{3}{2} \psi \left(-\dfrac{1}{2}\right) \)

Note that,

\(\displaystyle \psi \left(\dfrac{1}{2}\right) = -\gamma - \ln 4\)

\(\displaystyle \psi \left( \dfrac{3}{2} \right) = 2 - \gamma - \ln 4\)

\(\displaystyle \psi^{(1)} \left(\dfrac{3}{2}\right) = \dfrac{\pi ^2}{2} - 4 \)

\(\displaystyle \psi \left(-\dfrac{1}{2} \right) = 2 - \gamma - \ln 4\)

\(\displaystyle \implies \sum_{n=0}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = - \dfrac{\pi^2}{128}\)

\(\displaystyle \implies \sum_{n=1}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = \dfrac{1}{9} - \dfrac{\pi^2}{128} \) Ishan Singh · 9 months, 2 weeks ago

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@Ishan Singh Could someone please explain what \(\gamma\) is? I've seen it used in a lot of places without a definition. Is it a variable to represent infinity? Because the series \(\displaystyle\sum_{n=0}^{\infty} \frac{-1}{16(2n+1)}\) definitely diverges, for instance Ariel Gershon · 9 months, 2 weeks ago

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@Ariel Gershon @Ariel Gershon Note that the individual sums are not equal to the digamma values I wrote. They are assigned that value, i.e, if the individual diverging sums appear along with some other diverging series such that their value (taken together) is converging, then we can evaluate them by their assigned values. It is similar to assigning a value of \(-\dfrac{1}{12}\) to \(\zeta(-1)\), which in fact, diverges. \(\gamma\) is the Euler - Mascheroni constant which is equal to \(- \psi (1)\). Ishan Singh · 9 months, 1 week ago

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@Ariel Gershon It is euler-mascheroni constant. Aditya Kumar · 9 months, 2 weeks ago

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@Aditya Kumar Ok thanks Ariel Gershon · 9 months, 2 weeks ago

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@Ishan Singh Yeah even I did it the same way. I had posted a similar problem which uses digamma. Aditya Kumar · 9 months, 2 weeks ago

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@Ishan Singh nice proof! Hummus A · 9 months, 2 weeks ago

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@Hummus A @Hummus a How can we solve it using series expansion? (As you have stated) Samuel Jones · 9 months, 2 weeks ago

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@Samuel Jones and i now remember i used \({ x }^{ 2 }\sqrt { 1-{ x }^{ 2 } } ={ x }^{ 2 }-\displaystyle\sum _{ n=2 }^{ \infty }{ \frac { \begin{pmatrix} 2n-4 \\ n-2 \end{pmatrix} }{ (n-1){ 2 }^{ 2n-3 } } { x }^{ 2n } } \) Hummus A · 9 months, 2 weeks ago

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@Samuel Jones you can integrate its expansion,then let \(x=\sin{t}\) then integrate from 0 to pi/2(i think,it's some definite integral,and i'm not very organized with my work,i tend to do it at any piece of paper i have in front of me, yes i am very lazy),then do a lot of tedious arranging,then get it,it's the same technique i used in my second expansion of \(\pi^2\) Hummus A · 9 months, 2 weeks ago

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Here is a way to solve it without digamma functions:

We start out with the partial fraction expansion: \[128\sum_{n=1}^{\infty} \frac{n+1}{(2n+1)(2n-1)(2n+3)^2} = \sum_{n=1}^{\infty} \left(-\frac{8}{2n+1} + \frac{6}{2n-1} + \frac{2}{2n+3} - \frac{8}{(2n+3)^2}\right)\]

We can split this into three different sums: \[= \sum_{n=1}^{\infty} \left(\frac{8}{2n-1}-\frac{8}{2n+1}\right) - \sum_{n=1}^{\infty}\left(\frac{2}{2n-1} - \frac{2}{2n+3}\right) - \sum_{n=1}^{\infty}\frac{8}{(2n+3)^2}\]

The first sum is a telescoping sum: \[ \left(\frac{8}{1}- \color{red}{\frac{8}{3}}\right) + \left(\color{red}{\frac{8}{3}}-\color{green}{\frac{8}{5}}\right) + \left(\color{green}{\frac{8}{5}}-\frac{8}{7}\right) + ...\] Every term gets cancelled except the first one. Therefore the first sum is \(8\).

The second one is also a telescoping sum, but it's a little more tricky. The cancellation starts happening at the third term, so the result is the sum of the positive parts of the first two terms: \[\left(\frac{2}{1} - \color{red}{\frac{2}{5}}\right) + \left(\frac{2}{3} - \color{orange}{\frac{2}{7}}\right) + \left(\color{red}{\frac{2}{5}} - \color{green}{\frac{2}{9}}\right) + \left(\color{orange}{\frac{2}{7}} - \frac{2}{11}\right) + \left(\color{green}{\frac{2}{9}} - \frac{2}{13}\right) + ...\] Only the terms \(\frac{2}{1}+\frac{2}{3}\) don't get cancelled out. Therefore the value of the second sum is \(\frac{8}{3}\).

Now let's find the last sum, using the fact that \(\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}\): \[\sum_{n=1}^{\infty}\frac{8}{(2n+3)^2} = 8\left(\frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + ...\right)\]\[= 8\left[\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right) - \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + ...\right) - \frac{1}{1^2} - \frac{1}{3^2}\right]\]\[ = 8\left(\sum_{k=1}^{\infty} \frac{1}{k^2} - \frac{1}{2^2} \sum_{k=1}^{\infty}\frac{1}{k^2} - 1 - \frac{1}{9}\right)\]\[= 8\left(\frac{\pi^2}{6} - \frac{\pi^2}{4*6} - \frac{10}{9}\right) = \pi^2 - \frac{80}{9}\]

Therefore, the total is: \[\frac{128}{9} - \left[\sum_{n=1}^{\infty}\left(\frac{8}{2n-1}-\frac{8}{2n+1}\right) - \sum_{n=1}^{\infty}\left(\frac{2}{2n-1} - \frac{2}{2n+3}\right) - \sum_{n=1}^{\infty}\frac{8}{(2n+3)^2}\right]\]\[ = \frac{128}{9} - \left[8 - \frac{8}{3} - \left(\pi^2 - \frac{80}{9}\right)\right] = \pi^2 \] QED Ariel Gershon · 9 months, 2 weeks ago

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@Ariel Gershon

We can split this into three different sums: \[= \sum_{n=1}^{\infty} \left(\frac{8}{2n-1}-\frac{8}{2n+1}\right) - \sum_{n=1}^{\infty}\left(\frac{2}{2n-1} - \frac{2}{2n+3}\right) - \sum_{n=1}^{\infty}\frac{8}{(2n+3)^2}\]

Careful there, you need to justify why you can split an infinite sum into two or more infinite sums. Hint: Prove that all these infinite sums are absolutely convergent.

Otherwise, neat solution! =D =D Pi Han Goh · 9 months, 2 weeks ago

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@Pi Han Goh Good point. I did prove that each of the 3 sums converge, and since all the terms in each one are positive, they are absolutely convergent. :) Ariel Gershon · 9 months, 1 week ago

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@Pi Han Goh i totally agree Joel Yip · 9 months, 1 week ago

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