# Summation involving the integer nearest to $\sqrt n$.

Let $\langle n\rangle$ denote the integer nearest to $\sqrt n$. Evaluate $$\sum_{n=1}^{\infty} \frac{2^{\langle n\rangle}+2^{-\langle n\rangle}}{2^n}.$$

I tried writing down a few terms but I couldn't get any idea of how the sum progress. Any ideas? Thanks.

• Are you trying to decide convergence or actually compute the value? – Michael Burr May 4 '17 at 9:56
• I think "evaluate" is clear enough, but don't expect a closed expression ( unless you consider theta functions closed expressions). You could try to group together those $n$ with $<n>=k$. – Professor Vector May 4 '17 at 10:08

So now write down a few terms $$S=\frac{2^1+2^{-1}}{2^1}+\frac{2^1+2^{-1}}{2^2}+\frac{2^{2}+2^{-2}}{2^3} \text{so on}$$
Now consider the sequence $1,3,7,...$ The nth term of this sequence can be written as $n^2-n+1$ You can now convert the sum as $$\sum_{1}^{\infty} \frac{2^i+2^{-i}}{2^{i^2-i+1}}[1/2+1/2^2+...\text{2i terms}]$$ which after simplification van be shown to telescope yielding $3$ as answer.It can be seen from the attached picture.