Simplifying Repeated Infinite Summation

Question

While reading a solution to a math olympiad problem, I came across a repeated, infinite summation I've never seen before. The author somehow simplifies the summation to a numerical result, but doesn't elucidate how the simplification was made. It may involve a formula that I don't know of, but I am not sure. Could someone enlighten me with an explanation?

Here is a link to the full problem and solution. Thanks!

Answer 1

If $f(a)$, $g(b)$, and $h(c)$ are any functions, then a triple sum can be factored: $$ \sum_{a=0}^A \sum_{b=0}^B \sum_{c=0}^C f(a)g(b)h(c) = \bigg( \sum_{a=0}^A f(a)\bigg) \bigg( \sum_{b=0}^B g(b)\bigg) \bigg( \sum_{c=0}^C h(c) \bigg). $$ (This is just a souped-up version of $wy+wz+xy+xz=(w+x)(y+z)$.)

Under suitable conditions (which include $f$, $g$, and $h$ all being nonnegative), we can take the limit as $A,B,C\to\infty$ and obtain $$ \sum_{a=0}^\infty \sum_{b=0}^\infty \sum_{c=0}^\infty f(a)g(b)h(c) = \bigg( \sum_{a=0}^\infty f(a)\bigg) \bigg( \sum_{b=0}^\infty g(b)\bigg) \bigg( \sum_{c=0}^\infty h(c) \bigg). $$ This is what the author used, with $f(a)=1/2^a$ and so on.

Answer 2

Hint

The left hand side

$$=\sum_{a=0}^\infty\left(\dfrac12\right)^a\sum_{b=0}^\infty\left(\dfrac13\right)^b\sum_{c=0}^\infty\left(\dfrac15\right)^c$$

$$=\sum_{a=0}^\infty\left(\dfrac12\right)^a\sum_{b=0}^\infty\left(\dfrac13\right)^b\dfrac1{1-\dfrac15}$$