BdMO Online Forum

Problem 11:
Find $S$ where \

এর সমাধান কি ?
এখানে কি দুটো Sum দিয়ে সমষ্টির সমষ্টি বোঝাইছে ?? confused

i think it is summation of summation

যদিও সলুশন আমার অজানা, তবে এটা জানি যে সলুশনটা বেশ বড়।

Looks like I still have some gun powder left in my gun .
Sketch Solution:

Given \
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{\frac{3^m}{m}(\frac{3^m}{m}+\frac{3^n}{n})}\]
By symmetry \[S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{\frac{3^n}{n}(\frac{3^m}{m}+\frac{3^n}{n})} \]
So, \[2S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left (\dfrac{1}{\frac{3^n}{n}(\frac{3^m}{m}+\frac{3^n}{n})} + \dfrac{1}{\frac{3^m}{m}(\frac{3^m}{m}+\frac{3^n}{n})} \right )\]
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{ \frac{3^m}m +\frac {3^n}{n}} { \frac {3^m}{m} \frac {3^n}{n} (\frac{3^m}{m}+\frac{3^n}{n})}\]
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac {1} {\frac {3^m}{m} \frac {3^n}{n} }\]
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m}{3^m} \frac n{3^n}\]
\[ =\sum_{m=1}^{\infty} \frac{m}{3^m} \left ( \sum_{n=1}^{\infty} \frac n{3^n} \right ) \]
\[ = \left (\sum_{m=1}^{\infty} \frac{m}{3^m} \right) \left ( \sum_{n=1}^{\infty} \frac n{3^n} \right ) \]
Now, \[ \left (\sum_{m=1}^{\infty} \frac{m}{3^m} \right) = \frac 3 4 \]
So, \[2S = \frac 3 4 \times \frac 3 4 = \frac 9 {16}\]
And thus \
As far as I remember, the problem setter was Abir vai. He showed it in 2010 winter camp too.

[hide]Edit:
ভাত খাইতে গিয়া দেরি হয়া গেল দেখি -_-[/hide]

I found a straight solution without even symmetry, and I think this solutions makes more sense than the symmetry one. $S$ can be re-written as
\[\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2n}{3^n\cdot(n3^n+n3^n)}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2}{\left(3^n\right)^2}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \left(\dfrac{n}{3^n}\right)\left(\dfrac{n}{3^n}\right)=\dfrac12\left(\sum_{n=1}^\infty\dfrac n{3^n}\right)^2\]
Note the trick to write the sum as the last portion. It is actually the following:
\[\sum_{i=1}^k\sum_{j=1}^la_ib_j=\left(\sum_{i=1}^ka_i\right)\left(\sum_{j=1}^lb_i\right)\]
Now $T=\sum\limits_{n=1}^\infty\dfrac n{3^n}$ which can be found easily.

Masum wrote:I found a straight solution without even symmetry, and I think this solutions makes more sense than the symmetry one. $S$ can be re-written as
\[\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2n}{3^n\cdot(n3^n+n3^n)}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2}{\left(3^n\right)^2}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \left(\dfrac{n}{3^n}\right)\left(\dfrac{n}{3^n}\right)=\dfrac12\left(\sum_{n=1}^\infty\dfrac n{3^n}\right)^2\]
Note the trick to write the sum as the last portion. It is actually the following:
\[\sum_{i=1}^k\sum_{j=1}^la_ib_j=\left(\sum_{i=1}^ka_i\right)\left(\sum_{j=1}^lb_i\right)\]
Now $T=\sum\limits_{n=1}^\infty\dfrac n{3^n}$ which can be found easily.

I am not quite sure about this :/
For example, let $\{a_i \} = \{i\}$ and $\{b_i \} = \{\frac 1 i\}$.
Then according to your method \[\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \sum _{i=1}^{n} \sum _{i=1}^{n} a_ib_i = \sum _{i=1}^{n} \sum _{i=1}^{n} 1 = \sum _{i=1}^{n} \sum _{i=1}^{n} 1 \cdot 1 = \left (\sum _{i=1}^{n} 1 \right)^2 = n^2 \]
While \[\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \left( \sum _{i=1}^{n} i \right) \left( \sum _{i=1}^{n} \frac 1 i \right) \] which is definitely not $n^2$

Hmm. May be a miss-statement. This is infact a generalization of $a(c+d)+b(c+d)=(a+b)(c+d)$ so far as I am concerned.

Masum wrote:Hmm. May be a miss-statement. This is infact a generalization of $a(c+d)+b(c+d)=(a+b)(c+d)$ so far as I am concerned.

$\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = (\sum _{i=1}^{n} a_i) (\sum _{j=1}^{n} b_j)$ might be, but I am not sure about $\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \sum _{i=1}^{n} \sum _{i=1}^{n} a_ib_i $