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)

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Corollary: \(\sum_{n = 1}^{\infty} \frac{n}{a^n}\) for \(a > 1\) is equal to \(\frac{a}{(a-1)^2}\)

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 \(\frac{\frac{1}{a-1}}{1 - \frac{1}{a}} = \frac{a}{(a-1)^2}\).

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Thanks for the corollary. It will help many people I believe. :) I find just 1 thing confusing here.

Is \(S_n = \frac n{a^n}\)? If yes, why is \(S_2 = \frac 1{a(a-1)}\)?

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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.

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

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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 :)

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Hum, so so. The train of thought, at least for me, went like this "Oh, that is ugly. I don't like things that are not symetric. Can I factor it? No? Well, I can write as 1/(x^2+xy) each summand some x and y. What if I sum it to itself, reversed? That would make things symmetric. Oh, the summand is now 1/(xy) so it is solvable now."

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."

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

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Incidentally, one can also consider the series \(\frac{1}{1-x}=\sum_{n\geq0} x^n\) (convergent in (-1,1)) and differentiate it by \(x\), so we have \(\frac{1}{(1-x)^2}=\sum_{n\geq 0} n x^{n-1}\) so \(\frac{x}{(1-x)^2}=\sum_{n\geq 0} n x^{n}\). It is still convergent at \((-1,1)\) so we ca evaluate it in \(x=\frac{1}{3}\) and compute \(\sum_{n\geq 1} \frac{n}{3^n}=\frac{1/3}{(1-1/3)^2}=\frac{3}{4}\).

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)

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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.Log in to reply

Thanks a lot for the help. :)

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