π\pi!

π2=1289−128∑n=1∞n+1(2n+1)(2n−1)(2n+3)2 \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 t21−t2t^2\sqrt{1-t^2} and I came to the series above, can you prove it?

#Calculus

Note by Hamza A
5 years, 3 months ago

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Comments

Here is a quick method.

Using partial fractions, we have,

∑n=0∞n+1(2n+1)(2n−1)(2n+3)=∑n=0∞[−116(2n+1)+164(2n+3)−116(2n+3)2+364(2n−1)]\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,

=132ψ(12)−1128ψ(32)−164ψ(1)(32)+32ψ(−12)\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,

ψ(12)=−γ−ln⁡4\displaystyle \psi \left(\dfrac{1}{2}\right) = -\gamma - \ln 4

ψ(32)=2−γ−ln⁡4\displaystyle \psi \left( \dfrac{3}{2} \right) = 2 - \gamma - \ln 4

ψ(1)(32)=π22−4\displaystyle \psi^{(1)} \left(\dfrac{3}{2}\right) = \dfrac{\pi ^2}{2} - 4

ψ(−12)=2−γ−ln⁡4\displaystyle \psi \left(-\dfrac{1}{2} \right) = 2 - \gamma - \ln 4

  ⟹  ∑n=0∞n+1(2n+1)(2n−1)(2n+3)=−π2128\displaystyle \implies \sum_{n=0}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = - \dfrac{\pi^2}{128}

  ⟹  ∑n=1∞n+1(2n+1)(2n−1)(2n+3)=19−π2128\displaystyle \implies \sum_{n=1}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = \dfrac{1}{9} - \dfrac{\pi^2}{128}

Ishan Singh - 5 years, 3 months ago

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nice proof!

Hamza A - 5 years, 3 months ago

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@Hummus a How can we solve it using series expansion? (As you have stated)

Samuel Jones - 5 years, 3 months ago

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@Samuel Jones – you can integrate its expansion,then let x=sin⁡tx=\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 π2\pi^2

Hamza A - 5 years, 3 months ago

@Samuel Jones – and i now remember i used x21−x2=x2−∑n=2∞(2n−4n−2)(n−1)22n−3x2n{ 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 } }

Hamza A - 5 years, 3 months ago

Yeah even I did it the same way. I had posted a similar problem which uses digamma.

Aditya Kumar - 5 years, 3 months ago

Could someone please explain what γ\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 ∑n=0∞−116(2n+1)\displaystyle\sum_{n=0}^{\infty} \frac{-1}{16(2n+1)} definitely diverges, for instance

Ariel Gershon - 5 years, 3 months ago

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It is euler-mascheroni constant.

Aditya Kumar - 5 years, 3 months ago

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@Aditya Kumar – Ok thanks

Ariel Gershon - 5 years, 3 months ago

@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 −112-\dfrac{1}{12} to ζ(−1)\zeta(-1), which in fact, diverges. γ\gamma is the Euler - Mascheroni constant which is equal to −ψ(1)- \psi (1).

Ishan Singh - 5 years, 3 months ago

Here is a way to solve it without digamma functions:

We start out with the partial fraction expansion: 128∑n=1∞n+1(2n+1)(2n−1)(2n+3)2=∑n=1∞(−82n+1+62n−1+22n+3−8(2n+3)2)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: =∑n=1∞(82n−1−82n+1)−∑n=1∞(22n−1−22n+3)−∑n=1∞8(2n+3)2= \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: (81−83)+(83−85)+(85−87)+... \left(\frac{8}{1}- \color{#D61F06}{\frac{8}{3}}\right) + \left(\color{#D61F06}{\frac{8}{3}}-\color{#20A900}{\frac{8}{5}}\right) + \left(\color{#20A900}{\frac{8}{5}}-\frac{8}{7}\right) + ... Every term gets cancelled except the first one. Therefore the first sum is 88.

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: (21−25)+(23−27)+(25−29)+(27−211)+(29−213)+...\left(\frac{2}{1} - \color{#D61F06}{\frac{2}{5}}\right) + \left(\frac{2}{3} - \color{#EC7300}{\frac{2}{7}}\right) + \left(\color{#D61F06}{\frac{2}{5}} - \color{#20A900}{\frac{2}{9}}\right) + \left(\color{#EC7300}{\frac{2}{7}} - \frac{2}{11}\right) + \left(\color{#20A900}{\frac{2}{9}} - \frac{2}{13}\right) + ... Only the terms 21+23\frac{2}{1}+\frac{2}{3} don't get cancelled out. Therefore the value of the second sum is 83\frac{8}{3}.

Now let's find the last sum, using the fact that ∑k=1∞1k2=π26\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}: ∑n=1∞8(2n+3)2=8(152+172+192+1112+...)\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[(112+122+132+142+...)−(122+142+162+182+...)−112−132]= 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(∑k=1∞1k2−122∑k=1∞1k2−1−19) = 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(π26−π24∗6−109)=π2−809= 8\left(\frac{\pi^2}{6} - \frac{\pi^2}{4*6} - \frac{10}{9}\right) = \pi^2 - \frac{80}{9}

Therefore, the total is: 1289−[∑n=1∞(82n−1−82n+1)−∑n=1∞(22n−1−22n+3)−∑n=1∞8(2n+3)2]\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]=1289−[8−83−(π2−809)]=π2 = \frac{128}{9} - \left[8 - \frac{8}{3} - \left(\pi^2 - \frac{80}{9}\right)\right] = \pi^2 QED

Ariel Gershon - 5 years, 3 months ago

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We can split this into three different sums: =∑n=1∞(82n−1−82n+1)−∑n=1∞(22n−1−22n+3)−∑n=1∞8(2n+3)2= \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 - 5 years, 3 months ago

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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 - 5 years, 3 months ago

i totally agree

Joel Yip - 5 years, 3 months ago
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