fecundavai3c

2022-02-23

Does this double series has closed form (i.e. can be computed) ?
$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{10}^{nm}}$

Zernerqcw

Trivially
$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{10}^{mn}}=\sum _{m=1}^{\mathrm{\infty }}\frac{1}{{10}^{m}-1}=-\sum _{m=1}^{\mathrm{\infty }}\frac{1}{1-{10}^{m}}$
and the last sum can be computed, using the identity
$\sum _{m=1}^{\mathrm{\infty }}\frac{1}{1-{a}^{m}}=\frac{{\psi }_{\frac{1}{a}}\left(1\right)+\mathrm{log}\left(a-1\right)+\mathrm{log}\left(\frac{1}{a}\right)}{\mathrm{log}\left(a\right)}$
where
${\psi }_{q}\left(z\right)=-\mathrm{log}\left(1-q\right)+\mathrm{log}\left(q\right)\sum _{m=0}^{\mathrm{\infty }}\frac{{q}^{n+z}}{1-{q}^{n+z}}$
is the q-Polygamma function. Hence
$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{10}^{mn}}=-\frac{{\psi }_{\frac{1}{10}}\left(1\right)+\mathrm{log}\left(9\right)+\mathrm{log}\left(\frac{1}{10}\right)}{\mathrm{log}\left(10\right)}\approx 0.122324$

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