Finding ∑11+n2\sum \frac{1}{1+n^2}

How do we prove that the sum S=∑n=0∞11+n2S=\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} converges to π+12+πe2π−1\dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}?

#Calculus #InfiniteSeries

Note by Pratik Shastri
6 years, 8 months ago

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Comments

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

sin(x)=x∏n=1∞(1−x2(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=1∞ln(1−x2(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=1∞2x(nπ)21(x2(nπ)2−1)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=1∞2x(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+1π∑n=1∞2x(x2−n2)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+2x2∑n=1∞1(x2−n2)\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+2x2∑n=1∞1x2+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+2∑n=1∞1n2+1\pi coth(\pi )=1+2\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } }

=2∑n=0∞1n2+1−1=2\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } -1

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

⇒∑n=0∞1n2+1=π+12+πe2π−1\Rightarrow \displaystyle \sum_{n=0}^{\infty}{\frac{1}{n^{2}+1}}=\frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}

Ronak Agarwal - 6 years, 8 months ago

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

Pratik Shastri - 6 years, 8 months ago

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From where you found the question.

Ronak Agarwal - 6 years, 8 months ago

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@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 - 6 years, 8 months ago

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

Karan Siwach - 6 years, 8 months ago

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

Curtis Clement - 6 years, 5 months ago

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Easy, we know that :

cos(ix)=ex+e−x2cos(ix) = \frac{{e}^{x}+{e}^{-x}}{2}

Also sin(ix)=ex−ex2isin(ix) = \frac{{e}^{x}-{e}^{x}}{2i}

Dividing then we get :

cot(ix)=(ex+e−xex−e−x)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 - 6 years, 5 months ago

If somebody wants to overkill it, Then

S=∑n=0∞11+n2=12i∑n=0∞(1n−i−1n+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(1−i)2i 1+ \frac{\psi(1+i) - \psi(1-i)}{2i} Now, again using reflection formula, this can be easily calculated and it equals,

12+π2coth⁡π=π+12+πe2π−1 \frac{1}{2} + \frac{\pi}{2} \coth \pi = \frac{\pi+1}{2} + \frac{\pi}{e^{2\pi}-1}

Shivang Jindal - 6 years, 7 months ago

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

찬홍 민 - 6 years, 5 months ago
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