I need help with a double sum I need help with the following exercise:
determining the value of the next sum:
$$\sum _{ m=1 }^{ \infty  }{ \sum _{ n=1 }^{ \infty  }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) }  }  } $$
First time I face a double sum.
I have not the slightest idea how to begin to work it
 A: Let 
$$S = \sum _{ m=1 }^{ \infty  }{ \sum _{ n=1 }^{ \infty  }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) }  }  } $$
Using the fact that 
$$\sum_{m = 1}^\infty\sum_{n = 1}^\infty f(m,n) = \sum_{m = 1}^\infty\sum_{n = 1}^\infty f(n,m) $$
we can rewrite $S$ as
$$S = \sum _{ m=1 }^{ \infty  }{ \sum _{ n=1 }^{ \infty  }{ \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n{ 3 }^{ m }+m{ 3 }^{ n }) }  }  } $$
Summing up, we have
$$\begin{align}2S &= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left({{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) }  }  } + { \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n{ 3 }^{ m }+m{ 3 }^{ n }) }  }\right) \\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{3^nm^2n + 3^mn^2m}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn(m3^n + n3^m)}{3^{m+n}(n3^m + m3^n)}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{mn}{3^{m + n}}\\
&= \sum_{m = 1}^\infty\sum_{n = 1}^\infty \left(\frac{m}{3^m}\cdot\frac{n}{3^n}\right)\end{align}$$
But $\sum_{m = 1}^\infty\sum_{n = 1}^\infty x_my_n = \left(\sum_{m = 1}^\infty x_m \right)\left(\sum_{n = 1}^\infty y_n \right)$. Hence,
$$\begin{align}2S &= \left(\sum_{m = 1}^\infty \frac{m}{3^m}\right)\left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)\\
&= \left(\sum_{n = 1}^\infty \frac{n}{3^n}\right)^2\end{align}$$
To find this sum, we apply a classical trick: notice that for $|x| < 1$,
$$\frac{1}{1-x} = 1 + x + x^2 + \dots$$
Differentiating and multiplying both sides by $x$,
$$\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + \dots = \sum_{n = 1}^\infty nx^n$$
Substituting $x = \frac{1}{3}$,
$$\frac{3}{4} = \sum_{n = 1}^\infty \frac{n}{3^n}$$
So we deduce that
$$2S = \left(\frac{3}{4}\right)^2$$
and finally,
$$\sum _{ m=1 }^{ \infty  }{ \sum _{ n=1 }^{ \infty  }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) }  }  } = S = \frac{9}{32}$$
