TITLE: If $g\geq2$ is an integer, then $\sum\limits_{n=0}^{\infty} \frac{1}{g^{n^{2}}} $ and $ \sum\limits_{n=0}^{\infty} \frac{1}{g^{n!}}$ are irrational
QUESTION [7 upvotes]: How do we show that if $g \geq 2$ is an integer, then the two series $$\sum\limits_{n=0}^{\infty} \frac{1}{g^{n^{2}}} \quad  \ \text{and} \ \sum\limits_{n=0}^{\infty} \frac{1}{g^{n!}}$$ both converge to irrational numbers.
Well, i tried to see what happens, if they converge to a rational but couldn't get anything out it.

REPLY [9 votes]: Just for the record: the usual proof of irrationality is as follows (say for the first sum). Suppose that the sum is $p/q$. Multiply by $qg^{n^2}$, where $n$ is "big enough". We get that $$\sum_{m > n} \frac{q}{g^{m^2-n^2}}$$ is an integer. However, the latter is bounded by the geometric series $$q \sum_{t > 0} g^{-Nt} = \frac{q}{g^N} \cdot \frac{1}{1-g^{-N}} = \frac{q}{g^N-1},$$ where $N = (n+1)^2 - n^2 = 2n+1$. When $n$ is big enough, $q/(g^N-1) < 1$, contradiction.
This proofs works for any series $$\sum_{n=1}^\infty \frac{1}{\prod_{i=1}^n a_i}$$ where the non-zero integers $a_i$ satisfy $|a_i| \rightarrow \infty$.