The value of $\sum_{1\leq l< m How to solve this summation ? Also, I'm not sure what does $1\leq l< m <n$ supposed to imply in the development of summation form.
$$\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}$$
This one is from the Galois-Noether Contest in 2018:
Galois-Contest
 A: \begin{align*}\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\sum_{n=m+1}^\infty\frac{1}{2^n}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\sum_{m=l+1}^\infty \frac{1}{3^m}\cdot\frac{1}{2^m}\\&=\sum_{l=1}^\infty \frac{1}{5^l}\cdot\frac{1}{6^l}\cdot\frac15\\&=\left(\frac{1}{1-\frac1{30}}-1\right)\cdot\frac15\\&=\frac{1}{29\cdot5}\end{align*}
or rather $$\frac{1}{(5\cdot3\cdot2-1)(3\cdot2-1)(2-1)}$$
A: We have 
$$
\sum_{1\leq l<m<n}\frac{1}{5^l3^m2^n}=\sum_{l\geq1}\left(\frac{1}{5^l}\sum_{m>l}\left(\frac{1}{3^m}\sum_{n>m}\frac{1}{2^n}\right)\right).
$$
Now 
$$
\sum_{n=m+1}^\infty\frac{1}{2^n}=2^{-m},
$$
so we can reduce the sum to 
$$
\sum_{l\geq1}\left(\frac{1}{5^l}\sum_{m>l}\left(\frac{1}{6}\right)^m\right).
$$
Can you finish?
A: Outline:
You have
$$
\sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n}
= \sum_{n=1}^\infty\sum_{m=1}^{n-1}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell3^m2^n}
= \sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} \tag{1}
$$
Since $$
\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} = \frac{1}{5}\cdot\frac{1-1/5^{m-1}}{1-1/5} = \frac{1-1/5^{m-1}}{4} = \frac{1-5/5^{m}}{4}
$$
you can rewrite it as
$$
\sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n}
= \frac{1}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1-5/5^{m}}{3^m}
= \frac{1}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m}
- \frac{5}{4}\sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{15^m}\,.
$$
Since $\sum_{m=1}^{n-1}\frac{1}{3^m}$ and $\sum_{m=1}^{n-1}\frac{1}{15^m}$ can be computed as [...], you can rewrite [...]
Can you continue?
