A problem by Anish Roy

Squaring $(m-\frac{1}{2})$ we get $m^{2}-m+\frac{1}{4}$ . So $\sqrt{n}\geq m+\frac{1}{2}$ when $n \geq m^{2}-m+\frac{1}{4}$ , but $n$ is an integer, so $n \geq m^{2}-m+1$ . Squaring $(m+\frac{1}{2})$ we get $m^{2}+m+\frac{1}{4}$ so $\sqrt{n}< m+\frac{1}{2}$ when $n<m^{2}+m+\frac{1}{4}$ , but $n$ is an integer, so $n \leq m^{2}+m$ . Therefore, $\langle{n}\rangle=m$ , provided $n \in [m^{2}-m+1,m^{2}+m]$ . We can now rewrite our sum as: $\begin{aligned} \displaystyle \sum_{n=1}^{\infty} \frac {2^{\langle{n}\rangle}+2^{-\langle{n}\rangle}}{2^{n}} &= \displaystyle \sum_{m=1}^{\infty} \sum_{k=m^2-m+1}^{m^2+m} \frac {2^m+2^{-m}}{2^k} \\ & = \displaystyle \sum_{m=1}^{\infty} \frac {2^{2m}+1}{2^m} \sum_{k=m^2-m+1}^{m^2+m} \frac {1}{2k} \\ & = \displaystyle \sum_{m=1}^{\infty} \frac {2^{2m}+1}{2^{m+m^2-m+1}} \sum_{k=0}^{2m-1} \frac {1}{2k} \\ &= \displaystyle \sum_{m=1}^{\infty} \frac {2^{2m}+1}{2^{m^2+1}} \times \frac {1-2^{-2m}}{1-2^{-1}} \\ & = \displaystyle \sum_{m=1}^{\infty} \frac {2^{2m}+1}{2^{m^2+1}} \times \frac {2^{2m}-1}{2^{2m}-1} \\ & = \displaystyle \sum_{m=1}^{\infty} \frac {2^{4m}-1}{2^{m^2+2m}} \\ & = \displaystyle \sum_{m=1}^{\infty}(2^{1-(m-1)^2}-2^{1-(m-1)^2}) \\ & =2^1+2^0=3 \end{aligned}$ The last step follows from the fact that even and odd sub-sums telescope. So, just the first even and odd terms survive. So our answer is $\boxed{3}$