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# 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
3 years 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) · 3 years 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 $$\frac{\frac{1}{a-1}}{1 - \frac{1}{a}} = \frac{a}{(a-1)^2}$$. · 3 years 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 $$S_2 = \frac 1{a(a-1)}$$? · 3 years 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.

I hope this is helpful. · 3 years ago

got it, thanks :) · 2 years, 12 months ago

Is interchanging variables a common way to deal with such summations? · 3 years 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 :) · 3 years 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." · 3 years ago

You always want to "homogenize" the expression somehow because that often simplifies things. · 3 years 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) · 3 years ago

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

Hint: Interchange the order of summation. Staff · 3 years ago