Does this double sum $\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{m+n+mn}{2^m(2^m+2^n)}$ converge?

Question

$$\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{m+n+mn}{2^m(2^m+2^n)}$$ I have no experience evaluating double sums, but what intuition I have about single sums suggests to me that this series should converge. However, Mathematica fails to evaluate the sum and WolframAlpha tells me the series diverges.

I'm wondering if the averaging technique used here might yield something useful, but I don't if I'm adapting it properly. If $S$ denotes my sum, then $$2S=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{m+n+mn}{2^{m+n}}$$ which I'm told also diverges. I don't think this method will work since I can't split the numerator into a product $a_mb_n$.

Answer 1

Your initial series is convergent: you may consider partial sums to manipulate terms as you did, and since $$\displaystyle n+m+mn\leq 3(m+1)(n+1), \quad n\geq0,\,m\geq0,$$ you obtain $$ 0<2S=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{m+n+mn}{2^{m+n}}\leq\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{3(m+1)(n+1)}{2^{m+n}}=3\left(\sum_{n=0}^\infty \frac{n+1}{2^n}\right)^2=48. $$

Answer 2

Under assumption that it does converge, you can renumber the series. Imagine a $nm$ plane, and sample the points diagonally. In other words, say $k=m+n$ and write

$$2S=\sum_{k=0}^\infty\frac{1}{2^k} \sum_{n=0}^k (k+kn-n^2)$$ $$=\sum_{k=0}^\infty\frac{1}{2^k} \left(k(k+1)+k\frac{k(k+1)}{2}-\frac{k(k+1) (2k+1)}{6}\right)$$ $$=\frac16\sum_{k=0}^\infty\frac{1}{2^k} k(k+1)(k+5)$$ This is a convergent series: polynomial times a geometric progression has the same radius of convergence as the geometric series (for 1/2 in this case) and it can be evaluated by taking the derivative of the generating funcion $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$ and then setting $x=1/2$.

Answer 3

$$ \begin{align} \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m+n+mn}{2^{m+n}} &=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(m+1)(n+1)}{2^{m+n}} -\sum_{m=0}^\infty\sum_{n=0}^\infty\frac1{2^{m+n}}\\ &=\left(\sum_{n=0}^\infty\frac{n+1}{2^n}\right)^2 -\left(\sum_{n=0}^\infty\frac1{2^n}\right)^2\\[6pt] &=12 \end{align} $$

Answer 4

I'm not sure if your goal is specifically evaluating the double sum or just showing it converges (per your intuition), but we can show the sum converges as follows:

\begin{align} \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m+n+mn}{2^m(2^m+2^n)}&\leq\sum_{m=0}^\infty\left[\frac{1}{2^m}\sum_{n=0}^\infty\frac{m+n+mn}{2^n}\right]\\ &\leq\sum_{m=0}^\infty\left[\frac{m}{2^{m-1}}+\frac{C}{2^m}+\frac{mC}{2^m}\right], \end{align}

where $C$ denotes the convergent sum $\sum_{n=1}^\infty n/2^n$. The right-hand side is easily evaluated to give a finite number, hence the original sum converges.

Answer 5

$$S\le\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{m+n+mn}{2^m 2^n}=\sum_{m=0}^\infty\frac{m}{2^m} \sum_{n=0}^\infty \frac{1}{2^n}+\sum_{m=0}^\infty\frac{1}{2^m} \sum_{n=0}^\infty \frac{n}{2^n}+\sum_{m=0}^\infty\frac{m}{2^m} \sum_{n=0}^\infty \frac{n}{2^n}< \infty.$$