For any real number $m$, $ \left|\sum_{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $
Problem. Prove that for any real number $m$, $$ \left|\sum_{n=1}^{\infty}\frac{m}{n^2+m^2}\right|<\frac{\pi}{2} $$
Notes. This is an exercise in calculus. There are different ways to approach this problem. One may find the sum on the left explicitly with hyperbolic functions. One can also solve this problem by elementary estimates of integrals; I will write this one below.
1 answer
Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has $$ \frac{m}{n^2+m^2} = \int_{n-1}^{n}\frac{m}{n^2+m^2}\ dx <\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx $$ It follows that $$ \begin{align} \sum_{n=1}^\infty\frac{m}{n^2+m^2} &=\frac{m}{1^2+m^2}+\sum_{n=2}^\infty\frac{m}{n^2+m^2}\\ &<\int_{0}^{1}\frac{m}{x^2+m^2}\ dx +\sum_{n=2}^\infty\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx\\ &=\int_0^\infty\frac{m}{x^2+m^2}\ dx = \frac{\pi}{2} \end{align} $$ where the last integral is found by a change of variable $u=x/m$.
0 comment threads
0 comment threads