How to read and execute this sum?∑1≤l&lt;m&lt;n15l3m2n

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cindakayumn

2022-02-22

How to read and execute this sum?

\sum_{1 \leq l < m < n} \frac{1}{5^{l} 3^{m} 2^{n}}

an2gi2m9gg

Beginner2022-02-23Added 9 answers

Rewrite as triple sum

\sum_{1 \leq l < m < n} \frac{1}{5^{l} 3^{m} 2^{n}} = \sum_{l = 1}^{\infty} \sum_{m = l + 1}^{\infty} \sum_{n = m + 1}^{\infty} \frac{1}{5^{l} 3^{m} 2^{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}}

Then, this should be obvious in terms of how to read and execute.
Indeed, since each of them are Geometric Progressions, we have

\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}} (\frac{1}{2^{m}})

= \sum_{l = 1}^{\infty} \frac{1}{5^{l}} \sum_{m = l + 1}^{\infty} \frac{1}{6^{m}}

= \sum_{l = 1}^{\infty} \frac{1}{5^{l}} (\frac{1}{5 \cdot 6^{l}})

= \frac{1}{5} \sum_{l = 1}^{\infty} \frac{1}{30^{l}}

= \frac{1}{5} \cdot \frac{1}{29}

= \frac{1}{145}

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