# 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

## 1 Answer

Here is my solution

I bet you are preparing for indian statistical institute exam. (Im also preparing so best of luck)

First of all notice that there are 2i integers whose nearest integer to square root is i.It can be seen by taking a few examples. 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.

• Thank you.your solution is very good and you guessed right.best of luck to you too. – Navin May 4 '17 at 10:31
• – Shaun May 4 '17 at 11:09