Sequences and series (2)

$\large \displaystyle S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2 n}{3^m ( n \cdot 3^m + m \cdot 3^n ) }$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2 n}{3^m ( n \cdot 3^m + m \cdot 3^n ) } + \frac{mn^2}{3^n ( m \cdot 3^n + n \cdot 3^m ) } \right ]$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{ 3^n \cdot m^2n + 3^m \cdot mn^2}{3^{m+n}( n \cdot 3^m + m \cdot 3^n )} \right ]$

$\large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn ( n \cdot 3^m + m \cdot 3^n ) }{3^{m+n}( n \cdot 3^m + m \cdot 3^n )} \right ]$

$\large \displaystyle S = \frac12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn}{3^{m+n}}$

$\large \displaystyle S = \frac12 \sum_{m=1}^{\infty} \frac{m}{3^m} \sum_{n=1}^{\infty} \frac{n}{3^n}$

Note that $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ , for $|x| < 1$ . Differentiating w.r.t $x$ and multiplying by $x$ on both sides, one gets that $\sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2}$ .

So:

$\large \displaystyle S = \frac12 \left [ \frac{1/3}{(1-1/3)^2} \cdot \frac{1/3}{(1-1/3)^2} \right ]$

$\color{#20A900} \boxed{ \large \displaystyle S = \frac{9}{32} }$

Then:

$\color{#3D99F6} \large \displaystyle a = 9, b = 32, \boxed{ \large \displaystyle a+b=41}$

Replace 'm' with 'n' and 'n' with 'm' ,

Now add these two expressions,

Then we get terms which will be in A.G.P.,

So, by simplifying we get answer to be ${9}/{32}$ .

So $a+b$ = $41$ .