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

#NumberTheory #Symmetry #Series #Alzebra #BdMO

Easy Math Editor

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Comments

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)

Diego Roque - 7 years, 4 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 $\frac{\frac{1}{a-1}}{1 - \frac{1}{a}} = \frac{a}{(a-1)^2}$ .

Michael Tong - 7 years, 4 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 $S_2 = \frac 1{a(a-1)}$ ?

Labib Rashid - 7 years, 4 months ago

@Labib Rashid – 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.

I hope this is helpful.

Mursalin Habib - 7 years, 4 months ago

@Mursalin Habib – got it, thanks :)

Labib Rashid - 7 years, 4 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)

Diego Roque - 7 years, 4 months ago

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

Rahul Saha - 7 years, 4 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."

Diego Roque - 7 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 :)

Nur Muhammad Shafiullah - 7 years, 4 months ago

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

Michael Tong - 7 years, 4 months ago

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

Hint: Interchange the order of summation.

Calvin Lin Staff - 7 years, 4 months ago

Thanks a lot for the help. :)

Labib Rashid - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$