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# Finding $$\sum \frac{1}{1+n^2}$$

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

Note by Pratik Shastri
2 years ago

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

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

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

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

Differentiating both sides with respect to $$x$$ we get :

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

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

Putting $$\pi x$$ instead of $$x$$ we get :

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

Multiplying both sides with $$\pi x$$ we get :

$$\pi xcot(\pi x)=1+{ 2x }^{ 2 } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ ({ x }^{ 2 }-{ n }^{ 2 }) } }$$

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

Putting $$ix$$ in place of $$x$$ we get :

$$\pi xcoth(\pi x)=1+2{ x }^{ 2 } \displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+{ n }^{ 2 } } }$$

Put $$x=1$$ to get :

$$\pi coth(\pi )=1+2\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } }$$

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

$$\Rightarrow \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n^{ 2 }+{ 1 } } } =\frac { \pi coth(\pi )+1 }{ 2 }$$

$$\Rightarrow \displaystyle \sum_{n=0}^{\infty}{\frac{1}{n^{2}+1}}=\frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$$ · 2 years ago

That's some identity, isn't it? Nice solution by the way! · 2 years ago

From where you found the question. · 2 years ago

I didn't find it from anywhere. I was fiddling around with infinite series and I typed this into wolfram alpha. · 2 years ago

From where you get these type of identities. · 2 years ago

Can you please prove that $$icot(ix) = coth(x)$$ · 1 year, 9 months ago

Easy, we know that :

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

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

Dividing then we get :

$$cot(ix) = (\frac{{e}^{x}+{e}^{-x}}{{e}^{x}-{e}^{-x}})i$$

Hence finally we have :

$$cot(ix)=icoth(x)$$ · 1 year, 9 months ago

If somebody wants to overkill it, Then

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

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

$\frac{1}{2} + \frac{\pi}{2} \coth \pi = \frac{\pi+1}{2} + \frac{\pi}{e^{2\pi}-1}$ · 2 years ago

Are you 16 years old and know digamma function? Wow! · 1 year, 10 months ago