Expert answers to "Does this double series has closed form (i.e...."

fecundavai3c

Answered question

2022-02-23

Does this double series has closed form (i.e. can be computed) ?

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{1}{10^{n m}}

Zernerqcw

Beginner2022-02-24Added 11 answers

Trivially

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{1}{10^{m n}} = \sum_{m = 1}^{\infty} \frac{1}{10^{m} - 1} = - \sum_{m = 1}^{\infty} \frac{1}{1 - 10^{m}}

and the last sum can be computed, using the identity

\sum_{m = 1}^{\infty} \frac{1}{1 - a^{m}} = \frac{ψ_{\frac{1}{a}} (1) + \log (a - 1) + \log (\frac{1}{a})}{\log (a)}

where

ψ_{q} (z) = - \log (1 - q) + \log (q) \sum_{m = 0}^{\infty} \frac{q^{n + z}}{1 - q^{n + z}}

is the q-Polygamma function. Hence

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{1}{10^{m n}} = - \frac{ψ_{\frac{1}{10}} (1) + \log (9) + \log (\frac{1}{10})}{\log (10)} \approx 0.122324

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?