# Exact formula for $\sum_{n=0}^{+\infty}\frac1{n^2+1}$

Actually, as the corresponding integral $$\frac{\ln(\cos x)}{1-x}$$ or something like that) cannot be expressed in closed form by Liouville theorem, this shouldn't exist, but I believe I have seen it shown somewhere using Fourier series. By the way, the result : $$\sum_{n=0}^{\infty}\frac{1}{n^2+1}= \frac{1+ \pi\coth\pi}{2}$$ is known by WolframAlpha. But my question is : does there exist a general theory of similar results (something like exact formulas for $$\sum_{n=0}^{+\infty}\frac{1}{an^2+bn+c}$$, at least when $$a$$, $$b$$ and $$c$$ are integers)?

• Yes. We may interpret $f(n)$ as a residue of function $f(x)\cot (\pi x)$ in a point $n$. Then note that sum of residues of this function equals to 0 as integral over appropriate large circuit tends to 0. Nov 30, 2015 at 15:13
• There is no relation between existence of closed form for $\int f(x)\;dx$ and closed form for $\sum f(n)$. Nov 30, 2015 at 15:15
• Integral not of f(x), but of $f(x)\cot \pi x$ Nov 30, 2015 at 15:30

To do $\sum\frac{1}{an^2+bn+c}$, factor the denominator and get a digamma answer. $$\sum_{n=0}^\infty \frac{1}{(x+p)(x+q)} = \frac{\psi(p)-\psi(q)}{p-q}$$ And note that digamma of a rational can be evaluated in terms of logarithms and trig functions (Gauss's digamma theorem). $$\sum_{n=0}^\infty\frac{1}{(n+\frac{1}{4})(n+\frac{1}{3})} = 36\log 2+6\pi-2\pi\sqrt{3}-18\log 3 .$$

This is a standard exercise on residue theory. If $f$ is a rational function with zero of order $\geq 2$ at infinity and no poles at integers, then $$\sum_{-\infty}^\infty f(n)=-\sum{\mathrm{res}}_af(z)\pi\cot\pi z,$$ where the summation is over all poles of $f$. For $f(z)=1/(1+z^2)$ we obtain $$\sum_0^\infty\frac{1}{1+n^2}=1+\sum_{1}^\infty\frac{1}{1+n^2}=(\pi\coth\pi+1)/2,$$ so you copied the result from Wolfram incorrectly.

• Mmm ; sorry, but Wolfram (and Maple) gives the sum equals to 2.07417717505..., while (π coth π+1)/2 is 2.07667404... Nov 30, 2015 at 21:05
• @FeldmannDenis this anser 2.0766704 is correct, and when I ask Wolframalpha it confirms this wolframalpha.com/input/… Nov 30, 2015 at 22:02

Alternatively, you may use expansion of cotangent $$\pi\cot \pi x=\frac1x+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2}$$ as a blackbox (after all, complex residues are not the only way to obtain this formula: say, there is Herglotz trick, explained, for instance, in Proofs from the Book).

Just substitute $$x=i$$, for the sum $$S=\sum_{n=0}^{\infty} \frac1{n^2+1}$$ we get $$\pi \cot \pi i=-i-2i(S-1)$$, $$2iS=i-\pi \cot \pi i=i(1+\pi\coth \pi)$$, $$S=(1+\pi \coth \pi)/2$$.

For arbitrary rational function $$f(z)$$, we may find on this way the principal value of $$\sum_{n\in \mathbb{Z}} f(n)$$. If $$f(z)=\sum_k c_k/(z_k-z)$$, for each summand we have $$\sum_{j=-n}^n\frac1{z_k-j}=\frac1{z_k}+\sum_{k=1}^n\frac{2z_k}{z_k^2-j^2}\rightarrow \frac{\pi\cot \pi z_k}2,$$ thus principal value of $$\sum_{n\in \mathbb{Z}} f(n)$$ equals $$\sum_k c_k \frac{\pi\cot \pi z_k}2.$$ This works whenever $$f$$ has only simple poles and $$f(z)=O(1/z)$$ for large $$z$$. If additionally $$f(z)=O(1/z^2)$$, principal value is the same as ordinary sum. If $$f$$ has multiple poles, we may perturbate $$f$$ a bit and take a limit. Yes, this is one of approaches to residues.

If you need a sum not over integers, but over positive integers (it looks like it is the case), you instead of cotangent use the digamma $$\psi$$-function $$\psi(z)=-\gamma+\sum_{n=0}^{\infty} \left(\frac1{n+1}-\frac1{n+z}\right).$$ This is the essence of Gerald Edgard's answer.

• Yes, but he asked to sum $f(n)$ with a more general class of functions $f$. Dec 1, 2015 at 0:14
• Indeed, but this works for all rational functions. Dec 1, 2015 at 5:59
• Just to complete the picture, the expansion of the cotangent can be obtained by logarithmic differentiation from the infinite product for $\sin(\pi x)$, and the infinite product for $\sin(\pi x)$ can be obtained as dominated limit on the (easily computed) complete factorization of the polynomials $(1+iz/n)^n-(1-iz/n)^n$, that converge to $e^{iz}-e^{-iz}=2i\sin(z)$ (Is this Herglotz' proof?) Dec 1, 2015 at 11:24
• (no: I've just found Herglotz trick on line) Dec 1, 2015 at 11:32
• oh, I did not know this trick with polynomial Dec 1, 2015 at 12:07