Finding 11+n2\sum \frac{1}{1+n^2}

How do we prove that the sum S=n=011+n2S=\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} converges to π+12+πe2π1\dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}?

Note by Pratik Shastri
5 years 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

Euler while solving basel problem considered the function sin(x)sin(x) as an infinte product as :

sin(x)=xn=1(1x2(nπ)2)sin(x)=x\displaystyle \prod _{ n=1 }^{ \infty }{ (1-\dfrac { { x }^{ 2 } }{ { (n\pi ) }^{ 2 } } ) }

Taking ln()ln() both sides we get:

ln(sin(x))=ln(x)+n=1ln(1x2(nπ)2)ln(sin(x))=ln(x)+\displaystyle \sum _{ n=1 }^{ \infty }{ ln(1-\dfrac { { x }^{ 2 } }{ ({ n\pi })^{ 2 } } ) }

Differentiating both sides with respect to xx we get :

cot(x)=1x+n=12x(nπ)21(x2(nπ)21)cot(x)=\dfrac { 1 }{ x } +\displaystyle \sum _{ n=1 }^{ \infty }{ \dfrac { 2x }{ { (n\pi ) }^{ 2 } } \dfrac { 1 }{ (\dfrac { { x }^{ 2 } }{ { (n\pi ) }^{ 2 } } -1) } }

cot(x)=1x+n=12x(x2(nπ)2)\Rightarrow cot(x)=\dfrac { 1 }{ x } +\displaystyle \sum _{ n=1 }^{ \infty }{ \dfrac { 2x }{ ({ x }^{ 2 }-{ (n\pi ) }^{ 2 }) } }

Putting πx\pi x instead of xx we get :

cot(πx)=1πx+1πn=12x(x2n2)cot(\pi x)=\dfrac { 1 }{ \pi x } +\dfrac { 1 }{ \pi } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 2x }{ ({ x }^{ 2 }-{ n }^{ 2 }) } }

Multiplying both sides with πx\pi x we get :

πxcot(πx)=1+2x2n=11(x2n2)\pi xcot(\pi x)=1+{ 2x }^{ 2 } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ ({ x }^{ 2 }-{ n }^{ 2 }) } }

Also we know that icot(ix)=coth(x)icot(ix)=coth(x)

Putting ixix in place of xx we get :

πxcoth(πx)=1+2x2n=11x2+n2\pi xcoth(\pi x)=1+2{ x }^{ 2 } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+{ n }^{ 2 } } }

Put x=1x=1 to get :

πcoth(π)=1+2n=11n2+1\pi coth(\pi )=1+2\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } }

=2n=01n2+11=2\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } -1

n=01n2+1=πcoth(π)+12\Rightarrow \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } =\frac { \pi coth(\pi )+1 }{ 2 }

n=01n2+1=π+12+πe2π1\Rightarrow \displaystyle \sum_{n=0}^{\infty}{\frac{1}{n^{2}+1}}=\frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}

Ronak Agarwal - 5 years ago

Log in to reply

That's some identity, isn't it? Nice solution by the way!

Pratik Shastri - 5 years ago

Log in to reply

From where you found the question.

Ronak Agarwal - 5 years ago

Log in to reply

@Ronak Agarwal I didn't find it from anywhere. I was fiddling around with infinite series and I typed this into wolfram alpha.

Pratik Shastri - 5 years ago

Log in to reply

@Ronak Agarwal From where you get these type of identities.

Karan Siwach - 5 years ago

Log in to reply

Can you please prove that icot(ix)=coth(x)icot(ix) = coth(x)

Curtis Clement - 4 years, 9 months ago

Log in to reply

Easy, we know that :

cos(ix)=ex+ex2cos(ix) = \frac{{e}^{x}+{e}^{-x}}{2}

Also sin(ix)=exex2isin(ix) = \frac{{e}^{x}-{e}^{x}}{2i}

Dividing then we get :

cot(ix)=(ex+exexex)icot(ix) = (\frac{{e}^{x}+{e}^{-x}}{{e}^{x}-{e}^{-x}})i

Hence finally we have :

cot(ix)=icoth(x)cot(ix)=icoth(x)

Ronak Agarwal - 4 years, 9 months ago

Log in to reply

If somebody wants to overkill it, Then

S=n=011+n2=12in=0(1ni1n+i) S = \sum_{n=0}^{\infty} \frac{1}{1+n^2} \\ \\ = \frac{1}{2i} \sum_{n=0}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) Now , using the unique property of digamma function that is it satisfies, ψ(x+1)ψ(x)=1x \psi(x+1)-\psi(x)=\frac{1}{x}

We get that , sum is equivalent to, 1+ψ(1+i)ψ(1i)2i 1+ \frac{\psi(1+i) - \psi(1-i)}{2i} Now, again using reflection formula, this can be easily calculated and it equals,

12+π2cothπ=π+12+πe2π1 \frac{1}{2} + \frac{\pi}{2} \coth \pi = \frac{\pi+1}{2} + \frac{\pi}{e^{2\pi}-1}

Shivang Jindal - 4 years, 12 months ago

Log in to reply

Are you 16 years old and know digamma function? Wow!

찬홍 민 - 4 years, 9 months ago

Log in to reply

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...