If $m\geq2$ is an integer, then $\sum\limits_{n=1}^{\infty}m^{-n^2}$ is irrational Let $m  \geq2$ be an integer. I want to ask how to prove that the sum of the following series is irrational:
$$\sum _{n=1}^{\infty} \frac{1}{m^{n^2}}$$
 A: To get a grasp on what is going on here, we start with a simple, seemingly unrelated, problem, namely, we show that the decimal representation of $x=1/7$ is periodic. 
We know how to do that, right? One should perform the long division of $1$ by $7$ and very soon the digits produced by the division will begin to cycle. The result is $$x=0.14285714285714285714\ldots,$$ usually abbreviated as $x=0.\overline{142857}$. Thus, the decimal representation is indeed periodic... but now we might want to ask two questions: 


*

*What causes this periodic expansion?

*What does it mean?


Well, the meaning is clear: the expansion says that $$x=n\cdot10^{-6}+n\cdot10^{-12}+n\cdot10^{-18}+\cdots$$ with $n=142857$, that is, that $$x=n\cdot10^{-6}\cdot\sum\limits_{i=0}^{+\infty}10^{-6i}=\frac{n\cdot10^{-6}}{1-10^{-6}}=\frac{n}{10^6-1}.$$
In other words, we started from a representation of $x$ as a fraction, namely $x=1/7$, and we reached another representation of $x$ as a fraction, namely $x=142857/(10^6-1)$. If the goal is to compute the decimal representation of $x$, the second fraction is actually quite nice because, due to the simple fact that $142857\lt10^7$, the periodicity of the representation of $x$ becomes obvious. But we still have two mysteries to solve:


*

*Why did one obtain $10^6-1$ as the denominator?

*Why did one obtain $142857$ as the numerator?


At this point, one could note that, since the expression of every rational number as a reduced fraction is unique, one better have $10^6-1=7\cdot n$. In particular, $10^6-1$ should be a multiple of $7$, that is $10^6=1\pmod{7}$. Now, this rings a bell! One knows that, as soon as $k$ and $b$ are relatively prime, $k^{\phi(b)}=1\pmod{b}$. Since $10$ and $7$ are relatively prime and $\phi(7)=6$, indeed $10^6-1$ is a multiple of $7$, and the remark also explains the appearance of $6$ as the exponent of $10$. More importantly, it suggests a reason why the whole shebang holds and, at the same time, a way to vastly generalize our observations to any rational number.
So, we now consider any rational number $x=a/b$. We assume without loss of generality that $a\geqslant1$, $b\geqslant2$ (otherwise $x$ is an integer), $a\leqslant b-1$ (otherwise, shift $x$ by an integer), and that $b$ has no factor $2$ or $5$ (otherwise, multiply $b$ by powers of $5$ or $2$ to get a power of $10$, then multiplying $x$ by this power of $10$ simply shifts the expansion of $x$). Thus, $10$ and $b$ are relatively prime and $10^c=1\pmod{b}$, for some positive integer $c$. This means that $10^c=bd+1$ for some integer $d$, which implies that $x=n/(10^c-1)$ with $n=ad\lt10^c-1$. One gets $x=0.\overline{n_{c-1}n_{c-2}\cdots n_1}$, where $n=\sum\limits_{i=0}^{c-1}n_i\cdot10^i$ with $n_i$ in $\{0,1,\ldots,9\}$ is the decimal representation of $n$. This proves that the decimal representation of $x$ is indeed periodic and, at the same time, yields a way to compute this representation.
(When $a=1$ and $b=7$, indeed one gets $c=6$, $d=999999/7=142857$ and $n=142857$.)

Coming back finally to the question asked, one sees that there is nothing specific to the base $10$ here. Thus, consider any integer $m\geqslant2$ and some real number $x$ in $[0,1]$. Thus, $x=\sum\limits_{i=1}^{+\infty}x_im^{-i}$ for some sequence $(x_i)_{i\geqslant1}$ with values in $\{0,1,\ldots, m-1\}$. Then, $x$ is a rational number if and only if the sequence $(x_i)_{i\geqslant1}$ is ultimately periodic. 
In the case at hand, $x_i=1$ when $i$ is a square and $x_i=0$ otherwise, hence the gaps between the digits $1$ are unbounded. This forbids ultimate periodicity, hence $x$ is irrational.
