Dangerous Sigma

Let us check the first few terms of $\left \lfloor \sqrt n \right \rceil$ .

$\underbrace{1,1}_{\times 2},\underbrace{2,2,2,2}_{\times 4},\underbrace{3,3,3,3,3,3}_{\times 6},\underbrace{4,4,4,4,4,4,4,4}_{\times 8}, \cdots, \underbrace{k, k, \dots, k, k}_{\times 2k}, \cdots$

We note that the greatest $n$ such that $\left \lfloor \sqrt n \right \rceil = k$ is $\sum_{j=1}^k 2j = k^2 + k$ , then the smallest $n$ is $(k-1)^2 + (k-1)+1 = k^2 - k -1$ . That is for $\left \lfloor \sqrt n \right \rceil = k$ , $k^2 - k + 1 \le n \le k^2 + k$ (see proof below) and there are $2k$ of $n$ 's.

Then consider the summation over all $\lfloor \sqrt n \rceil = k$ :

$\begin{aligned} \sum_{n=k^2-k+1}^{k^2+k} \frac {2^{\left \lfloor \sqrt n \right \rceil} +2^{- \left \lfloor \sqrt n \right \rceil}}{2^n} & = \frac {2^k + 2^{-k}}{2^{k^2-k+1}} + \frac {2^k + 2^{-k}}{2^{k^2-k+2}} + \frac {2^k + 2^{-k}}{2^{k^2-k+3}} + \cdots + \frac {2^k + 2^{-k}}{2^{k^2+k}} \\ & = \frac {2^k + 2^{-k}}{2^{k^2-k+1}} \left(1 + \frac 12 + \frac 14 + \cdots + \frac 1{2^{2k-1}}\right) \\ & = \frac {2^k + 2^{-k}}{2^{k^2-k+1}} \left(\frac {1-\frac 1{2^{2k}}}{1-\frac 12}\right) = \frac {2^k + 2^{-k}}{2^{k^2-k}} \left(1-\frac 1{2^{2k}}\right) \\ & = \frac {2^{2k} + 1}{2^{k^2}} \left(\frac {2^{2k}-1}{2^{2k}}\right) = \frac {2^{4k}-1}{2^{k^2+2k}} = \frac 1{2^{k^2-2k}} - \frac 1{2^{k^2+2k}} \\ & = \frac 2{2^{k^2-2k+1}} - \frac 2{2^{k^2+2k+1}} = \frac 2{2^{(k-1)^2}} - \frac 2{2^{(k+1)^2}} \\ \implies \sum_{n=1}^\infty \frac {2^{\left \lfloor \sqrt n \right \rceil} +2^{- \left \lfloor \sqrt n \right \rceil}}{2^n} & = 2 \sum_{k=1}^\infty \left(\frac 1{2^{(k-1)^2}} - \frac 1{2^{(k+1)^2}} \right) = 2\left(1 + \frac 12\right) = \boxed 3 \end{aligned}$

---

Proof:

The following is using $\lfloor x \rceil = \left \lfloor x + \frac 12 \right \rfloor$ to prove that $\sqrt{k^2-k+1} + \frac 12 > k$ slightly, so that $\left \lfloor \sqrt{k^2-k+1} \right \rceil = \left \lfloor \sqrt{k^2-k+1} + \frac 12 \right \rfloor = k$ :

Assuming that $\sqrt{k^2-k+1} + \frac 12 > k$ is true, then $\sqrt{k^2-k+1} > k - \frac 12 \implies k^2 - k + 1 > k^2 - k + \frac 14$ . So it is in fact true. Now assuming that $\sqrt{k^2-k+1} + \frac 12 < k + 1$ is true, then $\sqrt{k^2-k+1} < k + \frac 12 \implies k^2 - k + 1 < k^2 + k + \frac 14$ . It is again true. Therefore, $k < \left \lfloor \sqrt{k^2-k+1} \right \rceil < k+1 \implies \left \lfloor \sqrt{k^2-k+1} \right \rceil = k$ . This means that $\left \lfloor \sqrt{(k+1)^2-(k+1)+1} \right \rceil = \left \lfloor \sqrt{k^2+k+1} \right \rceil = k+1$ and that $\left \lfloor \sqrt n \right \rceil = k$ , when $k^2 - k + 1 \le n \le k^2 + k$ .