A New Year, A New Hard Problem! 3 of 5

Thanks to @Syamand Hasam for the solution. Giving more details here.

$\begin{aligned} S & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {m^2n}{3^m\left(n3^m + m3^n\right)} \\ 2S & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {m^2n}{3^m\left(n3^m + m3^n\right)} + \sum_{n=1}^\infty \sum_{m=1}^\infty \frac {n^2m}{3^n\left(m3^n + n3^m\right)} \\ & = \sum_{m=1}^\infty \sum_{n=1}^\infty \left(\frac {m^2n}{3^m\left(n3^m + m3^n\right)} + \frac {n^2m}{3^n\left(m3^n + n3^m\right)} \right) \\ & = \sum_{m=1}^\infty \sum_{n=1}^\infty \left(\frac {3^nm^2n}{3^m3^n\left(n3^m + m3^n\right)} + \frac {3^mn^2m}{3^m3^n\left(n3^m + m3^n\right)} \right) \\ & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {3^nm^2n +3^mn^2m }{3^{m+n}\left(n3^m + m3^n\right)} \\ & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {mn\left(n3^m + m3^n\right)}{3^{m+n}\left(n3^m + m3^n\right)} \\ & = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac {mn}{3^{m+n}} \\ & = \left(\sum_{n=1}^\infty \frac n{3^n}\right)^2 \end{aligned}$

Now consider:

$\begin{aligned} S_1 & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{3^n} = \sum_{\color{#D61F06}n=0}^\infty \frac n{3^n} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{3^{n+1}} \\ & = \frac 13 \sum_{n=0}^\infty \frac n{3^n} + \frac 13 \sum_{n=0}^\infty \frac 1{3^n} \\ & = \frac 13 S_1 + \frac 13 \sum_{n=0}^\infty \frac 1{3^n} \\ & = \frac 12 \sum_{n=0}^\infty \frac 1{3^n} \\ & = \frac 12 \left(\frac 1{1-\frac 13}\right) \\ & = \frac 34 \end{aligned}$

Therefore, $S = \dfrac 12 \left(\dfrac 34\right)^2 = \dfrac 9{32} \approx \boxed{0.28}$ .

Let $\displaystyle f(m,n) = \frac{m^2n}{3^m(n3^m + m3^n)}$ , note that $\displaystyle f(m,n) + f(n,m) = \frac{mn}{n3^m + m3^n} \left(\frac{m}{3^m} + \frac{n}{3^n} \right) = \frac{mn}{3^m 3^n}$ . Now, letting $\displaystyle S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} f(m,n) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} f(n,m)$ , by re-arrangement, which is allowed here because the summands are positive. Therefore, $\displaystyle 2S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (f(m,n) + f(n,m)) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn}{3^m 3^n} = \left(\sum_{n=1}^{\infty}\frac{n}{3^n} \right)^2$ . Now we can obtain the sum via differentiating the following identity, $\displaystyle 1 + x + x^2 + \dots = \frac{1}{1-x}, \ (|x| < 1)$ , then multiplying by $x$ we obtain, $\displaystyle x + 2x^2 + 3x^3 + \dots = \frac{x}{(1-x)^2}, \ (|x| < 1)$ . Substitute $x = \frac{1}{3}$ , to obtain that, $\displaystyle \sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$ . Therefore,

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