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!
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 $2^{-3/4}$.