TITLE: Sum $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{10^{mn}}$
QUESTION [6 upvotes]: Does this double series has closed form (i.e. can be computed) ?
$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{10^{mn}}$$

REPLY [10 votes]: Trivially
$$\overset{\infty}{\underset{m=1}{\sum}}\overset{\infty}{\underset{n=1}{\sum}}\frac{1}{10^{mn}}=\overset{\infty}{\underset{m=1}{\sum}}\frac{1}{10^{m}-1}=-\overset{\infty}{\underset{m=1}{\sum}}\frac{1}{1-10^{m}}$$
 and the last sum can be computed, using the identity$$\overset{\infty}{\underset{m=1}{\sum}}\frac{1}{1-a^{m}}=\frac{\psi_{1/a}\left(1\right)+\log\left(a-1\right)+\log\left(1/a\right)}{\log\left(a\right)}$$
 where$$\psi_{q}\left(z\right)=-\log\left(1-q\right)+\log\left(q\right)\overset{\infty}{\underset{m=0}{\sum}}\frac{q^{n+z}}{1-q^{n+z}}
 $$
 is the q-Polygamma function. Hence$$\overset{\infty}{\underset{m=1}{\sum}}\overset{\infty}{\underset{n=1}{\sum}}\frac{1}{10^{mn}}=-\frac{\psi_{1/10}\left(1\right)+\log\left(9\right)+\log\left(1/10\right)}{\log\left(10\right)}\approx0.122324.$$