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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?

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    $\begingroup$ These types of series are usually very difficult to handle. Dr. Sonnhard Graubner relates that Mathematica says this sum can be handled as an Elliptic Theta Function. This may be the simplest form for this kind of sum. Note that $$\vartheta_2(u,q)=2q^{1/4}\sum_{n=0}^\infty q^{n^2+n}\cos((2n+1)u)$$ so your sum is easily seen to be $2^{-3/4}\vartheta_2(0,1/2)$. $\endgroup$ – robjohn Oct 27 '14 at 16:15
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(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.

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