jmc

Let $n$ and $k$ be positive integers such that $⟨n⟩=k.$ Then $$k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2},$$or $$k^2-k+\frac{1}{4}<n<k^2+k+\frac{1}{4}.$$Thus, for a given positive integer $k,$ the values of $n$ such that $⟨n⟩=k$ are $n=k^2-k+1,$ $k^2-k+2,$ $\dots ,$ $k^2+k.$ Thus, we can re-write the sum as $$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{2^{⟨n⟩}+2^{-⟨n⟩}}{2^n}& =\sum _{k=1}^{\infty }\sum _{n=k^2-k+1}^{k^2+k}\frac{2^{⟨n⟩}+2^{-⟨n⟩}}{2^n}\\ & =\sum _{k=1}^{\infty }(2^k+2^{-k})\left(\frac{1}{2^{k^2-k+1}}+\frac{1}{2^{k^2-k+2}}+\cdots +\frac{1}{2^{k^2+k}}\right)\\ & =\sum _{k=1}^{\infty }(2^k+2^{-k})\cdot \frac{2^{2k-1}+2^{2k-2}+\cdots +1}{2^{k^2+k}}\\ & =\sum _{k=1}^{\infty }(2^k+2^{-k})\cdot \frac{2^{2k}-1}{2^{k^2+k}}\\ & =\sum _{k=1}^{\infty }(2^{-k^2+2k}-2^{-k^2-2k})\\ & =(2^1-2^{-3})+(2^0-2^{-8})+(2^{-3}-2^{-15})+(2^{-8}-2^{-24})+\cdots \\ & =\boxed{3}.\end{array}$$