<s>[INST] 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
 [/INST] 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?
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