# Need help with - Bangladesh Mathematical Olympiad 2009 - Higher secondary level problem - 11

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 Note by Labib Rashid
5 years, 9 months ago

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We 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)

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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 $S_2 = \frac 1{a(a-1)}$?

- 5 years, 9 months ago

No, that's not what Michael meant.

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.

- 5 years, 9 months ago

got it, thanks :)

- 5 years, 9 months ago

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)

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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."

- 5 years, 9 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 :)

- 5 years, 9 months ago

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

- 5 years, 9 months ago

Hint: If the summation was separable in each variable, it would be much much easier.

Hint: Interchange the order of summation.

Staff - 5 years, 9 months ago

Thanks a lot for the help. :)

- 5 years, 9 months ago