Prove that the limit of a series, containing 1/{powers-of-2}, is not rational I have a series, $$x_n = \sum_{k=0}^n2^{-k^2-k}, \forall n \in N$$
I have to find it's limit and prove it is not in Q(it is not rational).
I tried to write it $x_n=1+\frac{1}{2^1*2^1}+\frac{1}{2^4* 2^2} + ... + \frac{1}{2^{n^2}*2^n}$ or $x_n=\sum_{k=0}^n\frac{1}{2^{k(k+1)}}$
I also observed that it might be convergent, because the general term converges to 0...
How should I proceed next?
I found out from some sources that I might get to a solution by writing down all the terms in base 2, but can't there be a different solution that does not involve writing the number in base 2??
Your help is very appreciated, so thank you!
 A: Too long for a comment. 
It is easy to check that your series is convergent by comparison test. One could compare $\{2^{-n^2-n}\}^{\infty}_{n=0}$ to $\{2^{-n}\}^{\infty}_{n=0}$ i.e.
$2^{-n^2-n}\leq2^{-n}$ for all $n\geq0$ therefore
$$\sum^{N}_{n=0}2^{-n^2-n}\leq\sum^{N}_{n=0}2^{-n}=\frac{1-2^{-(N+1)}}{1-2^{-1}}$$
and as $N\to\infty$ then 
$$\sum^{\infty}_{n=0}2^{-n^2-n}\leq\sum^{\infty}_{n=0}2^{-n}=\frac{1}{1-2^{-1}}=2<\infty$$
so the series converges.
Notice that the Jacobi theta function of second kind is defined as 
\begin{align}\nu_2(z,q)=\sum^{\infty}_{-\infty}q^{(n+1/2)^2}e^{(2n+1)iz}&=2\sum^{\infty}_{n=0}q^{(n+1/2)^2}\cos((2n+1)z)\\&=2q^{1/4}\sum^{\infty}_{n=0}q^{n^2+n}\cos((2n+1)z)\end{align}
In this particular case $q=1/2$ and $z=0$. Thus the limit of the series is equal to 
$$\sum^{\infty}_{n=0}2^{-n^2-n}=2^{-3/4}\nu_2(0,1/2)$$
Now $2^{-3/4}\notin\mathbb{Q}$ so it remains to check $\nu_2(0,1/2)$. I suspect it not to be rational neither a rational multiple of $2^{-3/4}$.
A: If it were rational, the binary (or any other base) expansion would have to be eventually periodic.  For this series this is clearly not the case, as the gaps between the 1s become larger and larger. 


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*'Eventually periodic' means that after some finite place in the expansion, the remaining part is periodic. 

