This is probably the hardest problem from that year. I cannot solve it. It would be great if anyone could help me with key ideas/full solution here. Even better if someone could post it at the source page where many are looking for a solution. Thanks in advance.

**Statement: **

Find \(S\) where \[S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^m(3^mn+3^nm)}\]

Problem source - BdMO forum - Problem 11

## Comments

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TopNewestWe have \[S=\sum_{m,n\geq 1}\frac{3^nm^2n}{3^{m+n}(3^mn+3^nm)}\] and exchanging variables \[S=\sum_{m,n\geq 1}\frac{3^mn^2m}{3^{n+m}(3^nm+3^mn)}\] Summing both we have \[2S=\sum_{m,n\geq 1}\frac{3^mn^2m+3^nm^2n}{3^{n+m}(3^nm+3^mn)}=\sum_{m,n\geq 1}\frac{mn(3^mn+3^nm)}{3^{n+m}(3^nm+3^mn)}\] and thus, factoring, \[2S=\sum_{m,n\geq 1}\frac{mn}{3^{n+m}}=\left(\sum_{n\geq 1}\frac{n}{3^{n}}\right)\left(\sum_{m\geq 1}\frac{m}{3^{m}}\right)\] which is a standard problem.

One way to solve that part is using \[\frac{3}{2}=\sum_{n\geq 0} \frac{1}{3^n}\]. Squaring and collecting terms with same denominator, \[\frac{9}{4}=\sum_{n\geq 0}\frac{n+1}{3^n}=3\sum_{n\geq 1} \frac{n}{3^n}\].

So we have \[2S= \left(\sum_{n\geq 1}\frac{n}{3^{n}}\right)\left(\sum_{m\geq 1}\frac{m}{3^{m}}\right)=\left(\frac{3}{4}\right)^2=\frac{9}{16}\]. Thus, \[S=\frac{9}{32}\].

(And I can do all that algebraic juggling without worring about convergence because all the terms are nonnegative, so any ordering would be always convergent to the same number or divergent) – Diego Roque · 3 years, 4 months ago

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Notice that this sum is equal to an infinite series of infinite series. Namely,

\(S_1 = \frac{1}{a} + \frac{1}{a^2} + \dots\)

\(S_2 = \frac{1}{a^2} + \frac{1}{a^3} + \dots\)

...

Which comes out to \(\frac{\frac{1}{a}}{1 - \frac{1}{a}} + \frac{\frac{1}{a^2}}{1 - \frac{1}{a}} + \dots\)

Which is yet another infinite series, equal to \(\frac{\frac{1}{a-1}}{1 - \frac{1}{a}} = \frac{a}{(a-1)^2}\). – Michael Tong · 3 years, 4 months ago

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Is \(S_n = \frac n{a^n}\)? If yes, why is \(S_2 = \frac 1{a(a-1)}\)? – Labib Rashid · 3 years, 4 months ago

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The first series is \(\frac{1}{a}+\frac{2}{a^2}+\frac{3}{a^3}+\cdots \).

How do we go about finding its sum? Notice that the first term of this series [\(\frac{1}{a}\)] is the first term of \(S_1\). The second term of this series [\(\frac{2}{a^2}\)] is the sum of the second term of \(S_1\) and the first term of \(S_2\). The third term of this series [\(\frac{3}{a^3}\)] is the sum of the third term of \(S_1\), the second term of \(S_2\) and the first term of \(S_3\). The fourth term of this series [\(\frac{4}{a^4}\)] is the sum of ... and onward.

So, if we want to find the sum of the original series, we have to find the sums of all the \(S_i\)'s. And it turns out that all the \(S_i\)'s are infinite series themselves. I think you can take it from here.

I hope this is helpful. – Mursalin Habib · 3 years, 4 months ago

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– Labib Rashid · 3 years, 4 months ago

got it, thanks :)Log in to reply

– Rahul Saha · 3 years, 4 months ago

Is interchanging variables a common way to deal with such summations?Log in to reply

– Nur Muhammad Shafiullah · 3 years, 4 months ago

Well, interchanging variables might seem a little more intuitive if you rewrite the summand as \[S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{\dfrac{3^m}{m}(\dfrac{3^m}{m}+\dfrac{3^n}{n})}\] and the rest of it would be quite similar to this one :)Log in to reply

Or "Basically that sum is ugly and not very symmeyric. If I sum it to itself, reversed, it will be symmetric. So I will do that." – Diego Roque · 3 years, 4 months ago

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– Michael Tong · 3 years, 4 months ago

You always want to "homogenize" the expression somehow because that often simplifies things.Log in to reply

Repeating this process of differentiating and multiplying by \(x\) we can compute any value of the kind \(\sum_{n\geq 0} n^k x^n\) for a fixed integer \(k\). And summing those sums we can compute \(\sum_{n\geq 0} P(n) x^n\) for any polynomial \(P\) and it all will be convergent in \((-1,1)\). (Or for any complex \(x\) inside the unit circle (\(|x|<1\)) but not necesarily at the border of the circle) – Diego Roque · 3 years, 4 months ago

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Hint:If the summation was separable in each variable, it would be much much easier.Hint:Interchange the order of summation. – Calvin Lin Staff · 3 years, 4 months agoLog in to reply

Thanks a lot for the help. :) – Labib Rashid · 3 years, 4 months ago

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