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!
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 $\dfrac{n}{a_n} \to 0$.