# How to find the sum of another series

How do I find the sum of a series

$$\sum_{1 \leq x < y < z}^\infty \frac{1}{3^x4^y5^z}$$

Write $$i=x,\,j=y-x,\,k=z-y$$ so your sum is $$\sum_{ijk}\frac{1}{60^i 20^j 5^k}$$, all indices starting at $$1$$. But $$\sum_i\frac{1}{n^i}=\frac{1}{n-1}$$, so the result is $$\frac{1}{59\times 19\times 4}$$.
Direct calculation through expansion: \begin{align}\sum_{1 \leq x < y < z}^\infty \frac{1}{3^x4^y5^z}= &\color{red}{\frac13\left[\frac1{4^2}\left(\frac1{5^3}+\frac1{5^4}+\cdots\right)+\frac1{4^3}\left(\frac1{5^4}+\frac1{5^5}+\cdots\right)+\cdots\right]}+\\ +&\color{blue}{\frac1{3^2}\left[\frac1{4^3}\left(\frac1{5^4}+\frac1{5^5}+\cdots\right)+\frac1{4^4}\left(\frac1{5^5}+\frac1{5^6}+\cdots\right)+\cdots\right]}+\cdots=\\ =&\color{red}{\frac13\left[\frac1{4^2}\cdot \frac{1}{4\cdot 5^2}+\frac1{4^3}\cdot \frac1{4\cdot 5^3}+\cdots\right]}+\\ +&\color{blue}{\frac1{3^2}\left[\frac1{4^3}\cdot \frac1{4\cdot 5^3}+\frac1{4^4}\cdot \frac1{4\cdot 5^4}+\cdots\right]}+\cdots=\\ =&\color{red}{\frac13\cdot \frac{1}{4^3\cdot 5^2}\cdot \frac{4\cdot 5}{19}}+\\ +&\color{blue}{\frac1{3^2}\cdot \frac{1}{4^4\cdot 5^3}\cdot \frac{4\cdot 5}{19}}+\cdots=\\ =&\frac{1}{3\cdot 4^2\cdot 5\cdot 19}\cdot \frac{3\cdot 4\cdot 5}{59}=\\ =&\frac1{4\cdot 19\cdot 59}.\end{align}