Question

Find sum of the series $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^{2}n}{3^{m}(n.3^{m} + m.3^{n})}$

Answer 1

$\frac{9}{32}$

Answer 2

Let the given series be $S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^{2}n}{3^{m}(n.3^{m} + m.3^{n})}$.

Let the general term be $T_{m,n} = \frac{m^{2}n}{3^{m}(n.3^{m} + m.3^{n})}$.

Consider the sum of $T_{m,n}$ and $T_{n,m}$:

$T_{m,n} + T_{n,m} = \frac{m^{2}n}{3^{m}(n.3^{m} + m.3^{n})} + \frac{n^{2}m}{3^{n}(m.3^{n} + n.3^{m})}$.

Let $D = n.3^{m} + m.3^{n}$.

$T_{m,n} + T_{n,m} = \frac{m^{2}n}{3^{m}D} + \frac{n^{2}m}{3^{n}D} = \frac{1}{D} \left( \frac{m^{2}n}{3^{m}} + \frac{n^{2}m}{3^{n}} \right)$.

Factor out $mn$:

$T_{m,n} + T_{n,m} = \frac{mn}{D} \left( \frac{m}{3^{m}} + \frac{n}{3^{n}} \right)$.

Substitute $D = n.3^{m} + m.3^{n}$ and combine the terms in the parenthesis:

$\frac{m}{3^{m}} + \frac{n}{3^{n}} = \frac{m.3^{n} + n.3^{m}}{3^{m}3^{n}}$.

So, $T_{m,n} + T_{n,m} = \frac{mn}{n.3^{m} + m.3^{n}} \cdot \frac{m.3^{n} + n.3^{m}}{3^{m}3^{n}} = \frac{mn}{3^{m}3^{n}} = \frac{mn}{3^{m+n}}$.

This identity holds for all $m,n \ge 1$.

Now, sum this identity over all $m,n \ge 1$:

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} (T_{m,n} + T_{n,m}) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{mn}{3^{m+n}}$.

The left side can be written as $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} T_{m,n} + \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} T_{n,m}$.

Let $S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} T_{m,n}$. The second term is also $S$ (by changing dummy variables).

So, the left side is $S+S = 2S$.

The right side is $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left(\frac{m}{3^m}\right) \left(\frac{n}{3^n}\right)$.

This is a product of two independent sums:

$\left( \sum_{m=1}^{\infty} \frac{m}{3^m} \right) \left( \sum_{n=1}^{\infty} \frac{n}{3^n} \right)$.

Let $S_1 = \sum_{k=1}^{\infty} \frac{k}{3^k}$.

We use the known formula for the sum of the series $\sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$.

For $x = 1/3$:

$S_1 = \frac{1/3}{(1-1/3)^2} = \frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$.

So, the right side is $S_1 \cdot S_1 = \left(\frac{3}{4}\right) \left(\frac{3}{4}\right) = \frac{9}{16}$.

Equating the two sides:

$2S = \frac{9}{16}$.

$S = \frac{9}{32}$.