π\pi!

π2=1289128n=1n+1(2n+1)(2n1)(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 t21t2t^2\sqrt{1-t^2} and I came to the series above, can you prove it?

Note by Hamza A
3 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Here is a quick method.

Using partial fractions, we have,

n=0n+1(2n+1)(2n1)(2n+3)=n=0[116(2n+1)+164(2n+3)116(2n+3)2+364(2n1)]\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)=γln4\displaystyle \psi \left(\dfrac{1}{2}\right) = -\gamma - \ln 4

ψ(32)=2γln4\displaystyle \psi \left( \dfrac{3}{2} \right) = 2 - \gamma - \ln 4

ψ(1)(32)=π224\displaystyle \psi^{(1)} \left(\dfrac{3}{2}\right) = \dfrac{\pi ^2}{2} - 4

ψ(12)=2γln4\displaystyle \psi \left(-\dfrac{1}{2} \right) = 2 - \gamma - \ln 4

    n=0n+1(2n+1)(2n1)(2n+3)=π2128\displaystyle \implies \sum_{n=0}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = - \dfrac{\pi^2}{128}

    n=1n+1(2n+1)(2n1)(2n+3)=19π2128\displaystyle \implies \sum_{n=1}^{\infty} \dfrac{n+1}{(2n+1)(2n-1)(2n+3)} = \dfrac{1}{9} - \dfrac{\pi^2}{128}

Ishan Singh - 3 years, 9 months ago

Log in to reply

nice proof!

Hamza A - 3 years, 9 months ago

Log in to reply

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

Samuel Jones - 3 years, 9 months ago

Log in to reply

@Samuel Jones you can integrate its expansion,then let x=sintx=\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 - 3 years, 9 months ago

Log in to reply

@Samuel Jones and i now remember i used x21x2=x2n=2(2n4n2)(n1)22n3x2n{ 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 - 3 years, 9 months ago

Log in to reply

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

Aditya Kumar - 3 years, 9 months ago

Log in to reply

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=0116(2n+1)\displaystyle\sum_{n=0}^{\infty} \frac{-1}{16(2n+1)} definitely diverges, for instance

Ariel Gershon - 3 years, 9 months ago

Log in to reply

It is euler-mascheroni constant.

Aditya Kumar - 3 years, 9 months ago

Log in to reply

@Aditya Kumar Ok thanks

Ariel Gershon - 3 years, 9 months ago

Log in to reply

@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 - 3 years, 9 months ago

Log in to reply

Here is a way to solve it without digamma functions:

We start out with the partial fraction expansion: 128n=1n+1(2n+1)(2n1)(2n+3)2=n=1(82n+1+62n1+22n+38(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(82n182n+1)n=1(22n122n+3)n=18(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: (8183)+(8385)+(8587)+... \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: (2125)+(2327)+(2529)+(27211)+(29213)+...\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=11k2=π26\sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}: n=18(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+...)112132]= 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=11k2122k=11k2119) = 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(π26π246109)=π2809= 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(82n182n+1)n=1(22n122n+3)n=18(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[883(π2809)]=π2 = \frac{128}{9} - \left[8 - \frac{8}{3} - \left(\pi^2 - \frac{80}{9}\right)\right] = \pi^2 QED

Ariel Gershon - 3 years, 9 months ago

Log in to reply

We can split this into three different sums: =n=1(82n182n+1)n=1(22n122n+3)n=18(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 - 3 years, 9 months ago

Log in to reply

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 - 3 years, 9 months ago

Log in to reply

i totally agree

Joel Yip - 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...