Value of a series How can I find the value of the series $$\sum_{n=0}^\infty\frac{1}{2^{n^2+n}}$$
I have proved it is convergent, using the comparison test but I am stuck when it comes to actually finding an exact value. Any help would be greatly appreciated.
My original problem actually goes like this. I am given a sequence x_n=$$\sum_{k=0}^n\frac{1}{2^{k^2+k}}$$. I have to prove that the rationals are not complete by making use of this sequence. Can you guys find me a more elementary proof now?  
 A: (1). Theorem. For $x\in \mathbb Q$ there exists $r\in \mathbb Q^+$ such that  $$(a,b \in Z \land  b\ne 0 \land x\ne a/b)\implies |x-a/b|>r/|b|.$$
Proof: Let $x=c/d$ with $c,d\in \mathbb Z$  . We have $$(x\ne a/b\land |x-a/b|\leq r/|b|)\iff$$ $$\iff 0<|x-a/b|<r/|b|\iff $$ $$\iff0<|c/d-a/b|<r/|b| \iff$$ $$\iff 0<|cb-ad|<r|d|.$$ But $cb-ad\in \mathbb Z$, so $0<|cb-ad|<r|d|\iff 1\leq |cb-ad|<r|d|.$  And we cannot have $1<r|d|$ when $r=1/2|d|.$
(2). For integer $n\geq 0$ let $x_n=\sum_{j=0}^n 2^{-j^2-j}.$ We can readily confirm that $(x_n)_n$ is a Cauchy  sequence. 
Assume, by contradiction, that $(x_n)_ n$ converges to $x\in \mathbb Q.$
We have $0<x-x_n<4\cdot 2^{-(n+1)(n+2)}$ for every $n.$ Because, by contradiction,   if $x-x_{n_0}\geq 4\cdot 2^{-(n_0+1)(n_0+2)}$ for some $n_0$ then for every $n>n_0$ we have $$x-x_n=(x-x_{n_0})+(x_{n_0}-x_n)=$$    $$=(x-x_{n_0})-\sum_{j=1+n_0}^n2^{-j^2-j}\geq$$    $$\geq (x-x_{n_0})-\sum_{j=1+n_0}^n2^{-(n_0+1)(n_0+2)}\;2^{-(n-n_0-1)}>$$     $$>(x-x_{n_0})-2\cdot 2^{-(n_0+1)(n_0+2)}\geq$$ $$
\geq 2\cdot 2^{(n_0+1)(n_0+2)}$$ which implies that the sequence  $(x_n)_{n>n_0}$ does not converge to $x.$ 
Now we have $x_n=A_n/B_n$ where $A_n\in \mathbb N$ and $B_n=2^{-n^2-n}$. We have $$ 0<x-A_n/B_n=x-x_n<4\cdot 2^{-(n+1)(n+2)}=2^{-2n}/B_n.$$ But for every $r\in \mathbb Q^+$ there exists $n\in \mathbb N$ such that $r<2^{-2n}$, so the Theorem is contradicted. So assuming that the Cauchy sequence $(x_n)_n$ converges to a rational is untenable.
Notes. (i). I tried to work in $\mathbb Q$ without assuming the existence of irrationals, nor of least upper bounds. (ii). The Theorem is  usually stated as follows : If $x\in \mathbb R $ and for every $r>0$ there exist integers $a,b$ such that $0<|x-a/b|<r/|b|$, then $x$ is irrational.
