Symmetric Sums are not very popular!

Let $S = \sum_{m,n \geq 1} \frac{m^2n}{2^m(m2^n+n2^m)}$

We observe that the given sum is symmetric that means if we swap the variables $m$ and $n$ , the answer would still remain the same. Thus, we can state that

$S = \sum_{m,n \geq 1} \frac{m^2n}{2^m(m2^n+n2^m)} = \sum_{n,m \geq 1} \frac{n^2m}{2^n(m2^n+n2^m)}$

Adding the two sums, we get

$\begin{aligned} 2S & = & \sum_{m,n \geq 1} \frac{m^2n}{2^m(m2^n+n2^m)} + \frac{n^2m}{2^n(m2^n+n2^m)} \\ & = & \sum_{m,n \geq 1} \frac{mn}{m2^n+n2^m} \left(\frac{m}{2^m}+\frac{n}{2^n}\right) \\ & = & \sum_{m,n \geq 1} \frac{mn}{m2^n+n2^m} \left(\frac{m2^n+n2^m}{2^m2^n}\right) \\ & = & \sum_{m,n \geq 1} \frac{mn}{2^m2^n} \\ & = & \left(\sum_{m \geq 1} \frac{m}{2^m}\right)^2 \\ \end{aligned}$

Now, using Geometric Expansion, we see that $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ Differentiating the above series w.r.t $x$ and then multiplying by $x$ , we get $\frac{x}{(1-x)^2} = \sum_{n=0}^\infty nx^n$

By setting $x = \frac{1}{2}$ , we get the above required sum.

Thus, $2S = \left(\frac{\frac{1}{2}}{(1-\frac{1}{2})^2}\right)^2 = 4$

$\boxed{S = 2}$

$\textbf{EDIT : }$ If you had a different approach then please do share it with us.