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 $a_mb_n$.

some brute force testing suggested that the series does indeed converge: to 6 — Dleep, Aug 07 '15 at 17:29

score 14 · Answer 1 · answered Aug 07 '15 at 17:35

14

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

answered Aug 07 '15 at 17:35

robjohn

345,667

I always seem to forget that I can add and subtract... Thanks! – user170231 Aug 07 '15 at 18:49
it should be $6$ not $12$ – Oussama Boussif Aug 07 '15 at 19:27
@OussamaBoussif robjohn was merely evaluating $2S$, not $S$. – user170231 Aug 07 '15 at 19:51
Ah yes I didn't see that oops ^^ – Oussama Boussif Aug 07 '15 at 19:58

Olivier Oloa · Accepted Answer · 2015-08-07T17:54:34.173

6

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 $$ 0<2S=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{m+n+mn}{2^{m+n}}\leq\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{3(m+1)(n+1)}{2^{m+n}}=3\left(\sum_{n=0}^\infty \frac{n+1}{2^n}\right)^2=48. $$

edited Aug 07 '15 at 17:54

answered Aug 07 '15 at 17:27

Olivier Oloa

120,989

1

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 $12$. – robjohn Aug 07 '15 at 17:31
@robjohn I will edit, thanks. – Olivier Oloa Aug 07 '15 at 17:36
@OlivierOloa Just curious as to the relevance of the $3$. Do we not have $n+m+nm\le (n+1)(m+1)=n+m+nm+1$? – Mark Viola Aug 07 '15 at 22:08
@Dr.MV Sure. But I did not see it first... Then I didn't want to change it, since it was about convergence :) Thanks. – Olivier Oloa Aug 07 '15 at 23:19
1

No worry. Just curious. – Mark Viola Aug 08 '15 at 00:19

score 4 · Answer 3 · answered Aug 07 '15 at 17:37

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.

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. — user170231, Aug 07 '15 at 18:08

orion · Answer 4 · 2015-08-07T17:34:16.813

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 $x=1/2$.

user 1591719 · Answer 5 · 2015-08-08T09:30:45.297

3

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

edited Aug 08 '15 at 09:30

answered Aug 08 '15 at 08:03

user 1591719

44,216
12
105
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When $m=0$ or $n=0$ you have $mn=0$ and not necessarily $m+n+mn \leq 3mn$. – Olivier Oloa Aug 08 '15 at 08:58
1

@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. – user 1591719 Aug 08 '15 at 09:25

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

5 Answers5