1
$\begingroup$

Can anyone help with this:

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

$\endgroup$

closed as off-topic by user21820, George Law, Claude Leibovici, Did, Nosrati Aug 25 '17 at 10:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, George Law, Claude Leibovici, Did, Nosrati
If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I'm curious what theorem proves it to be transcendental? $\endgroup$ – Erick Wong Jan 19 '15 at 17:12
  • $\begingroup$ I think I've seen a proof, based on closeness to rationals, that the number (in decimal) .101001000100001000001... is transcendental; I assumed that the same would apply to the number at hand, which in base 2 is .10001000001000000010000000001... $\endgroup$ – Mark Fischler Jan 19 '15 at 20:15
3
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.