Hardy and Wright irrational sums A few days ago, I read a proof from Hardy and Wright saying if $a_n$ is an integer sequence and if
$$\lim_{n\to \infty}\frac n{a_n}=0$$ 
then 
$$\sum_{n=0}^\infty10^{-a_n}$$
is irrational.
Unfortunately, I can't find the proof again. If someone knows the proof or knows where to find it I would be very grateful.
Thanks in advance!
 A: Here is a proof
I just whipped up.
Show:
If
$a_n$
is a positive increasing
sequence of integers and
$\dfrac{n}{a_n} 
\to 0
$
and
$b \ge 2$
is an integer
then
$S
=\sum_{n=1}^{\infty} b^{-a_n}
$
is irrational.
First,
since $a_n \ge n$,
$\sum_{n=1}^{\infty} b^{-a_n}
$
converges.
Let
$S_m
=\sum_{n=1}^{m} b^{-a_n}
$
and
$T_m
=\sum_{n=m+1}^{\infty} b^{-a_n}
$.
$S_m
=\sum_{n=1}^{m} b^{-a_n}
=b^{-a_m}\sum_{n=1}^{m} b^{a_m-a_n}
=\dfrac{\sum_{n=1}^{m} b^{a_m-a_n}}{b^{a_m}}
=\dfrac{s_m}{b^{a_m}}
$.
$\begin{array}\\
T_m
&=\sum_{n=m+1}^{\infty} b^{-a_n}\\
&=b^{-a_{m+1}}\sum_{n=m+1}^{\infty} b^{a_{m+1}-a_n}\\
&\le b^{-a_{m+1}}\sum_{n=m+1}^{\infty} b^{m+1-n}\\
&= b^{-a_{m+1}}\sum_{n=0}^{\infty} b^{-n}\\
&= \dfrac1{(1-1/b)b^{a_{m+1}}}\\
&\le \dfrac{2}{b^{a_{m+1}}}\\
\end{array}
$
Therefore
$\begin{array}\\
|S-S_m|
&=|S-\dfrac{s_m}{b^{a_m}}|\\
&=T_m\\
&\le \dfrac{2}{b^{a_{m+1}}}\\
&= \dfrac{s_m}{b^{a_{m}}}\dfrac{2}{s_m(b^{a_{m+1}-a_m})}\\
&= S_m\dfrac{2}{s_m(b^{a_{m+1}-a_m})}\\
\end{array}
$
If $S$ is rational,
then
$b^{a_{m+1}-a_m}
$
is bounded,
so
$a_{m+1}-a_m
$
is bounded.
If 
$a_{m+1}-a_m
\le c$
for some $c > 0$,
then
$a_{m+k}-a_m
\le ck$
or
$a_{m+k}
\le a_m+ck$
or
$\dfrac{a_{m+k}}{m+k}
\le \dfrac{a_m+ck}{m+k}$
or
$\begin{array}\\
\dfrac{m+k}{a_{m+k}}
&\ge \dfrac{m+k}{a_m+ck}\\
&= \dfrac{m+k}{a_m+ck}-\dfrac{1}{c}+\dfrac{1}{c}\\
&= \dfrac{c(m+k)-(a_m+ck)}{a_m+ck}+\dfrac{1}{c}\\
&= \dfrac{cm-a_m}{a_m+ck}+\dfrac{1}{c}\\
&\ge \dfrac{1}{2c}\\
\end{array}
$
for
large enough $k$
which contradicts
$\dfrac{n}{a_n}
\to 0$.
