I am trying to calculate the (double) sum $\sum_{i>0} \sum_{j>0} \frac{i^2j}{3^i(j3^i+i3^j)}$, and I think i got it, but can someone verify it and see if any parts can be improved? (Simplified, justified, etc.)
My solution:
$$S=\sum_{i>0} \sum_{j>0} \frac{i^2j}{3^i(j3^i+i3^j)}$$
$$=\sum_{i>0}\sum_{j>0}\frac{i}{3^i}\cdot \frac{ij}{j3^i+i3^j}$$
$$=\sum_{i>0}\sum_{j>0}\frac{i}{3^i}\cdot\frac{\frac{i}{3^i}\frac{j}{3^j}}{\frac{j}{3^j}+\frac{i}{3^i}}$$
Now notice
$$S=\sum_{i>0}\sum_{j>0}\frac{i}{3^i}\cdot\frac{\frac{i}{3^i}\frac{j}{3^j}}{\frac{j}{3^j}+\frac{i}{3^i}}=\sum_{i>0}\sum_{j>0}\frac{j}{3^j}\cdot\frac{\frac{j}{3^j}\frac{i}{3^i}}{\frac{i}{3^i}+\frac{j}{3^j}}=S$$
$$\therefore S=\frac{1}{2}\sum_{i>0}\sum_{j>0}(\frac{i}{3^i}+\frac{j}{3^j})\cdot\frac{\frac{i}{3^i}\frac{j}{3^j}}{\frac{j}{3^j}+\frac{i}{3^i}}$$
$$=\frac{1}{2}\left(\sum_{i\geq 1}\frac{i}{3^i}\right)\left(\sum_{j\geq 1}\frac{j}{3^j}\right)$$
$$=\frac{1}{2}\left(\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\cdots\right)\left(\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\cdots\right)$$
Let $T=\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\cdots$
$$\frac{1}{3}T=\frac{1}{9}+\frac{2}{27}+\frac{3}{81}+\cdots$$
$$\frac{2}{3}T=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$$
$$T=\frac{3}{4}$$
$$\therefore S=\frac{1}{2}T^2=\frac{9}{32}$$
Thank you for helping.