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$$\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{m+n+mn}{2^m(2^m+2^n)}$$ 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 $a_mb_n$.

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  • $\begingroup$ some brute force testing suggested that the series does indeed converge: to 6 $\endgroup$ – Dleep Aug 7 '15 at 17:29
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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 $$ 0<2S=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{m+n+mn}{2^{m+n}}\leq\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{3(m+1)(n+1)}{2^{m+n}}=3\left(\sum_{n=0}^\infty \frac{n+1}{2^n}\right)^2=48. $$

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    $\begingroup$ Don't we have to worry about when $m=0$ or $n=0$ where $m+n+mn\ge 3mn$? It's not too hard, but not really trivial. In fact, the original sum is actually $12$. $\endgroup$ – robjohn Aug 7 '15 at 17:31
  • $\begingroup$ @robjohn I will edit, thanks. $\endgroup$ – Olivier Oloa Aug 7 '15 at 17:36
  • $\begingroup$ @OlivierOloa Just curious as to the relevance of the $3$. Do we not have $n+m+nm\le (n+1)(m+1)=n+m+nm+1$? $\endgroup$ – Mark Viola Aug 7 '15 at 22:08
  • $\begingroup$ @Dr.MV Sure. But I did not see it first... Then I didn't want to change it, since it was about convergence :) Thanks. $\endgroup$ – Olivier Oloa Aug 7 '15 at 23:19
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    $\begingroup$ No worry. Just curious. $\endgroup$ – Mark Viola Aug 8 '15 at 0:19
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$$ \begin{align} \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m+n+mn}{2^{m+n}} &=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(m+1)(n+1)}{2^{m+n}} -\sum_{m=0}^\infty\sum_{n=0}^\infty\frac1{2^{m+n}}\\ &=\left(\sum_{n=0}^\infty\frac{n+1}{2^n}\right)^2 -\left(\sum_{n=0}^\infty\frac1{2^n}\right)^2\\[6pt] &=12 \end{align} $$

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  • $\begingroup$ I always seem to forget that I can add and subtract... Thanks! $\endgroup$ – user170231 Aug 7 '15 at 18:49
  • $\begingroup$ it should be $6$ not $12$ $\endgroup$ – Oussama Boussif Aug 7 '15 at 19:27
  • $\begingroup$ @OussamaBoussif robjohn was merely evaluating $2S$, not $S$. $\endgroup$ – user170231 Aug 7 '15 at 19:51
  • $\begingroup$ Ah yes I didn't see that oops ^^ $\endgroup$ – Oussama Boussif Aug 7 '15 at 19:58
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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.

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  • $\begingroup$ I'm concerned with finding out if WA's conclusion is mistaken, so just telling whether or not the sum converges would suffice. Judging by robjohn's answer, finding the actual sum is a good exercise. $\endgroup$ – user170231 Aug 7 '15 at 18:08
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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 $x=1/2$.

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

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    $\begingroup$ When $m=0$ or $n=0$ you have $mn=0$ and not necessarily $m+n+mn \leq 3mn$. $\endgroup$ – Olivier Oloa Aug 8 '15 at 8:58
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    $\begingroup$ @OlivierOloa Yes, you're right! I typed that in a hurry. Still, I think the best way to show it is to remove $2^m$ from the brackets. $\endgroup$ – user 1357113 Aug 8 '15 at 9:25

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