jmc

The sum is taken over all positive integers $n$ and $k$ such that $k\le n-1,$ or $n\ge k+1.$ Thus, we can change the order of summation: $$\begin{array}{rl}\sum _{n=2}^{\infty }\sum _{k=1}^{n-1}\frac{k}{2^{n+k}}& =\sum _{k=1}^{\infty }\sum _{n=k+1}^{\infty }\frac{k}{2^{n+k}}\\ & =\sum _{k=1}^{\infty }\frac{k}{2^k}\sum _{n=k+1}^{\infty }\frac{1}{2^n}\\ & =\sum _{k=1}^{\infty }\frac{k}{2^k}\left(\frac{1}{2^{k+1}}+\frac{1}{2^{k+2}}+\cdots \right)\\ & =\sum _{k=1}^{\infty }\frac{k}{2^k}\cdot \frac{1}{2^k}\\ & =\sum _{k=1}^{\infty }\frac{k}{4^k}.\end{array}$$Let $$S=\sum _{k=1}^{\infty }\frac{k}{4^k}=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\cdots .$$Then $$4S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{3^3}+\cdots .$$Subtracting these equations, we get $$3S=1+\frac{1}{4}+\frac{1}{4^2}+\cdots =\frac{4}{3},$$so $S=\boxed{\frac{4}{9}}.$