TITLE: Triple summation verification $\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$
QUESTION [6 upvotes]: I am attempting the following triple summation. It would be great if someone would verify whether what I've done is correct.

$$\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}$$


$$\begin{aligned}\sum_{1\le i\lt j\lt k}\frac{1}{2^i3^j5^k}&=\sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\sum_{k=j+1}^{\infty}\frac{1}{2^i3^j5^k}\\&^= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\frac{1}{2^i3^j}\frac{1/5^{j+1}}{4/5}\\&=\frac{1}{4}\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{1/15^{i+1}}{14/15}\\&=\frac{1}{56}\sum_{i=1}^{\infty}\frac{1}{30^{i}}=\frac{1}{56}\cdot\frac{1}{29}\end{aligned}$$

REPLY [3 votes]: Your answer is correct but there is a easier (to remember) way to derive the result.
To simplify the sum, the first thing one should do is get rid of the "$<$" constraint among the summation indices. If you introduce index $\ell, m$ such that $j = i + \ell$ and $k = j + m$, then
the constraint $1 \le i < j < k$ simplifies to $1 \le i,\ell,m$. This will
allow you to rewrite your iterated sum as a product of geometric series:
$$\begin{align}\sum_{\substack{(i,j,k)\\1 \le i < j < k}}\frac{1}{2^i 3^j 5^k}
&= \sum_{\substack{(i,j,k)\\1 \le i, \ell, m }}\frac{1}{2^i 3^{i+\ell} 5^{i+\ell + m}}
=\sum_{i=1}^\infty \frac{1}{30^i}\sum_{\ell=1}^\infty\frac{1}{15^\ell}\sum_{m=1}^\infty\frac{1}{5^m}\\
&= \frac{\frac{1}{30}}{1-\frac1{30}}\frac{\frac{1}{15}}{1-\frac{1}{15}} \frac{\frac15}{1-\frac15} = \frac{1}{29}\frac{1}{14}\frac{1}{4} = \frac{1}{1624}
\end{align}
$$