How to find the sum $\sum_{i=1}^{\infty} \frac{1}{n^2+1}$

Asked Sep 22 '17 at 09:34

Active May 11 '19 at 11:00

Viewed 126 times

Given that this particular series is convergent by the integral test $$ \sum_{i=1}^{\infty} \frac{1}{n^2+1}$$

how can one find the value to which it converges?

Integral test:

\begin{align*} \int_1^{\infty} \frac{1}{x^2+1} \ dx = \lim_{a \to \infty} \int_{1}^{a} \frac{1}{x^2+1} = \lim_{a \to \infty} \left[ \arctan(a) - \arctan(1) \right]= \frac{\pi}{4} \end{align*}

edited May 11 '19 at 11:00

StubbornAtom

17,052

asked Sep 22 '17 at 09:34

bru1987

2,147
5
25
50

It's also convergent by the comparison test: It's smaller than $\sum1/n^2 = \pi^2/6$. – Arthur Sep 22 '17 at 09:34
@Arthur I see thatyou have approached the "convergence test" in a different way. But is that a way to actually define, precisely, the value of that sum? (for example as a fraction, say 4/3) – bru1987 Sep 22 '17 at 09:39
1

See this. – David Mitra Sep 22 '17 at 09:39
1

Good old wolfie alpha says $\frac12\left(\pi\coth(\pi) - 1\right)$. – Arthur Sep 22 '17 at 09:40
1

@Arthur Sure it follows from the partial fraction decomposition of trigonometric functions, eg. of $\frac{\pi^2}{\sin^2(\pi z)}$ – reuns Sep 22 '17 at 09:41
Thank you for the inputs, I see that it is not trivial at all! – bru1987 Sep 22 '17 at 09:43
$$\coth z=\frac{1}{z}+2z\sum_{n=1}^{\infty}\frac{1}{z^{2}+n^{2}\pi^{2}}$$ is formula 4.36.3 in standard tables. – Sep 22 '17 at 09:47

How to find the sum $\sum_{i=1}^{\infty} \frac{1}{n^2+1}$

0 Answers0