$\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{mn^2+m^2n+2mn}$

Question

Evaluate : $$\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{mn^2+m^2n+2mn}$$

My try :

Some of the terms are getting cancelled out but the rest are not

Answer 1

Here is an idea that can help: \begin{align*} \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{1}{mn^2+m^2n+2mn}& = \sum_{n=1}^\infty \frac{1}{n}\sum_{m=1}^\infty \frac{1}{m(m+n+2)}\\ & = \sum_{n=1}^\infty \frac{1}{n(n+2)}\sum_{m=1}^\infty \left[\frac{1}{m}-\frac{1}{m+n+2}\right]\\ & = \sum_{n=1}^\infty \frac{1}{n(n+2)}\left[\left(1-\frac{1}{n+3}\right)+\left(\frac{1}{2}-\frac{1}{n+4}\right)+\dotsb\right]\\ &\text{the expression inside the inner parentheses is actually a $\color{red}{\text{finite}}$ sum}\\ & = \sum_{n=1}^\infty \frac{1}{n(n+2)}\left[1+\frac{1}{2}+\dotsb +\frac{1}{n+2}\right]\\ & =\frac{1}{2} \sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+2}\right)\left[1+\frac{1}{2}+\dotsb +\frac{1}{n+2}\right]\\ & =\frac{1}{2} \left[\left(\frac{1}{1}-\frac{1}{3}\right)\left(1+\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\dotsb\right]\\ &=\frac{1}{2}\left[\left(1+\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\color{red}{\frac{1}{3}\left(\frac{1}{4}+\frac{1}{5}\right)+\frac{1}{4}\left(\frac{1}{5}+\frac{1}{6}\right)+\dotsb}\right]\\ &=\frac{1}{2}\left[\frac{11}{6}+\frac{25}{24}+\color{red}{\frac{1}{3}\left(\frac{1}{4}+\frac{1}{5}\right)+\frac{1}{4}\left(\frac{1}{5}+\frac{1}{6}\right)+\dotsb}\right]\\ &=\frac{1}{2}\left[\frac{11}{6}+\frac{25}{24}+\color{red}{\left(\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+ \dotsb\right)}+\color{blue}{\left(\frac{1}{3 \cdot 5}+\frac{1}{4 \cdot 6}\dotsb \right)}\right] \end{align*} Observe that the series in red and blue are telescoping series. That can be summed as $$\sum_{k=3}^{\infty}\frac{1}{k(k+a)}=\frac{1}{a}\sum_{k=3}^{\infty}\left(\frac{1}{k}-\frac{1}{k+a}\right)$$ Hopefully now you can finish it.