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

Question

Can anyone help with this:

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

Answer 1

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.

Answer 2

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.