<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] \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 $$\frac{1}{(5\cdot3\cdot2-1)(3\cdot2-1)(2-1)}$$
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