Double summation

We have:

$\displaystyle I= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn^2}{2^n(n2^m+m2^n)}$

$\displaystyle = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn^2}{4^n2^m(\frac{n}{2^n}+\frac{m}{2^m})}$ (Divide by $2^m2^n$ ).

Now:

$\displaystyle 2I= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn^2}{4^n2^m(\frac{n}{2^n}+\frac{m}{2^m})} + \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{nm^2}{4^m2^n(\frac{m}{2^m}+\frac{n}{2^n})}$

$\displaystyle = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn^2}{4^n2^m(\frac{n}{2^n}+\frac{m}{2^m})}+\frac{nm^2}{4^m2^n(\frac{m}{2^m}+\frac{n}{2^n})}$

$\displaystyle = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{\frac{mn}{2^m2^n}}{\frac{m}{2^m}+\frac{n}{2^n}} (\frac{n}{2^n}+\frac{m}{2^m})$

$\displaystyle = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn}{2^m2^n}$

$\displaystyle = (\sum_{m=1}^{\infty} \frac{m}{2^m} )^2$

The last series can be derived from the geometric series :

$\displaystyle \sum_{k=1}^{\infty} x^k = \frac{x}{1-x}$ where $x\in ]-1,1[$

Differentiate both sides:

$\displaystyle \sum_{k=1}^{\infty} kx^{k-1} = \frac{1}{(1-x)^2}$

Multiply by $x$ :

$\boxed{\displaystyle \sum_{k=1}^{\infty} kx^k = \frac{x}{(1-x)^2} }$

Therefore:

$\displaystyle 2I= (\sum_{m=1}^{\infty} m(\frac{1}{2})^m )^2 = (\frac{\frac{1}{2}}{(1-\frac{1}{2})^2})^2 = 4$

Hence, $\boxed{I=2}$ .