# Analytical solution of a convergent series [duplicate]

Which is the analytical solution of the infinite sum $\frac{1}{1+1^{2}}+\frac{1}{1+2^{2}}+\frac{1}{1+3^{2}}+\frac{1}{1+4^{2}}+\cdots$?

• Is it an university test? – Exodd Apr 11 '18 at 20:46
• What's an "analytical solution of an infinite sum"? (And why does your question say "Which" rather than "What"?) Do you want to know the limit of your series? Or what? – Rob Arthan Apr 11 '18 at 20:48
• It seems that you want to know the value of the summation in closed-form, i.e., a closed-form expression of the limit of the partial sums. Although mathematicians are very smart guys, they will not understand (or pretend to not understand) what you are asking until you use the right words... – rafa11111 Apr 11 '18 at 20:50
• Anyway, you might want to check Wolfram Alpha (wolframalpha.com/input/?i=sum+1%2F(1%2Bn%5E2)+from+1+to+infty ), that gives $(\pi \ \mathrm{cotanh} \ \pi - 1)/2$ as result. – rafa11111 Apr 11 '18 at 20:53
• How we get this result? – billmark Apr 11 '18 at 21:04

We start with this identity for the sine function: $$\frac{\sinh x}{x} = \prod_{k=1}^\infty \left(1+\frac{x^2}{k^2\pi^2}\right)$$
For details on its derivation, check this article. This product formula was already used to prove another interesting identity involving $\pi$ and infinite series!
Well, we also know that $$\frac{d}{dx} \log \sinh x = \coth x.$$
Due to the properties of the logarithm function, $\log \sinh x$ is $$\log x + \sum_{k=1}^\infty \log \left(1+\frac{x^2}{k^2\pi^2}\right),$$ therefore, $$\coth x = \frac{1}{x} + 2x \sum_{k=1}^\infty \frac{1}{k^2 \pi^2 + x^2}.$$
Using $x = \pi$ leads to $$\coth \pi = \frac{1}{\pi} + \frac{2}{\pi} \sum_{k=1}^\infty \frac{1}{k^2 + 1},$$ then, solving for the required summation, $$\sum_{k=1}^\infty \frac{1}{k^2 + 1} = \frac{\pi \coth \pi - 1}{2}.$$