Is $\sum_{n=1}^\infty \frac{1}{2^{n^2}}$ is rational number? Can anyone help with this:

Is $\sum_{n=1}^{\infty}\frac{1}{2^{n^2}}$ is rational number?

 A: It isn't. It's quite obvious to prove if you have ever seen the simple proof why $e$ is an irrational number.
Assume that the sum is equal to $p / q$, for some integers $p$ and $q$.
Multiply the sum by $q \cdot 2^{k^2}$, then the result must be an integer. Now pick $k$ large enough that $2^{2k+1}$ is much bigger than $q$.
In the sum, the first $k$ terms multiplied by $q \cdot 2^{k^2}$ are integers. But the next term $1 / 2^{(k+1)^2}$ multiplied by $q \cdot 2^{k^2}$ is $q / 2^{2k+1}$ which is much smaller than 1, and the following terms are even smaller, so they don't add up to $1$.
So the sum multiplied by $q \cdot 2^{k^2}$ is not an integer. Since $q$ was an arbitrary integer, the sum is not rational. 
A: No, in fact, 
$$S = \sum_1^\infty 2^{-n^2}$$
is not only irrational, it is transcendental.
I will prove here that it is irrational:
Suppose $S$ is rational, then write it as a reduced fraction $ S = \frac{p}{q}$ with 
$\text{gcd} (p,q) = 1$. Let $k = \lfloor \log_2 q \rfloor$. Now define
$$S_i = \sum_1^i 2^{-n^2} = \sum_1^i T_n$$
so that for any $i$, 
$$S = S_i + \sum_{n=i+1}^\infty T_n$$
It is easy to show that 
$$\sum_{n=i+1}^\infty T_n = 2^{-(i+1)^2} \left(1 + \sum_{n=i+2}^\infty 2^{-n^2}\right)
< 2^{-(i+1)^2} \left(1 + \sum_{n=(i+2)^2}^\infty 2^{-n}\right) =
 2^{-(i+1)^2} \left(1 + 2^{-i^2-4i-3}\right)\\
\sum_{n=i+1}^\infty T_n<  2^{-(i+1)^2} + 2^{-2i^2-6i-4}< 2^{-i^2+1}
$$
If we choose $i > k+1$, then 
$$S_i = \frac{w}{2^{k+1}}$$ with $e$ an integer; so 
$S$ is between  $\frac{w}{2^{k+1}}$ and $\frac{w+ {\epsilon}}{2^{k+1}}$ and we can take $i$ large enough that the difference which contradicts the assumption about $S$ being $p/q$.
Sorry this proof got sloppy toward the end.
