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

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.

Solution 1:

Both are transcendental numbers. This follows from Roth's theorem on the rate of approximations to irrational numbers by rationals. See also this question. The key concept is "irrationality measure", see for example this entry from MathWorld.

Solution 2:

The numbers are irrational, as they are infinite non-repeated "decimal" numbers in base-$g$.

Solution 3:

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 $|a_i| \rightarrow \infty$.