TITLE: Infinite Triple Sum
QUESTION [5 upvotes]: The following triple sum appeared in a recent mathematical competition at a local university:
$$\sum_{j,k,l\geq0} \frac{1}{3^l\left(3^{j+k}+3^{k+l}+3^{l+j}\right)}$$
I know the answer is $\frac{9}{8}$, but I have no clue how to proceed with the problem.
As an addendum, what is the general closed form of the sum when $3$ is replaced with a positive integer other than $1$?

REPLY [6 votes]: Let $$S=\sum_{i,j,k \geq 0} \frac{1}{3^i(3^{i+j}+3^{i+k}+3^{j+k})}$$
Note that $$\frac{1}{3^i}+\frac{1}{3^j}+\frac{1}{3^k} = \frac{3^{i+j}+3^{i+k}+3^{j+k}}{3^{i+j+k}}$$
Hence $$3S=\sum_{i,j,k \geq 0} \frac{1}{3^{i+j+k}} = (\frac{3}{2})^3$$
therefore $S=9/8$.