Show that
$$\sum_{n=1}^{\infty} \dfrac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n} = 3$$
I think I can do the following which I am not quite sure about:
$$\sum_{n=1}^{\infty} \dfrac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n} =\sum_{n=1}^{\infty}\dfrac{1}{2^{n-\sqrt{n}}}+\dfrac{1}{2^{n+\sqrt{n}}}$$
But after this I am quite not sure how to form the two infinite GP series. Please help.
I have also learned that $$\sum_{n=1}^{[x]} a^{\sqrt{n}} = \dfrac{2}{(\log_e{a})^2}+c$$ for $a>0$ and $\in \mathbb{N}$
Does this help my problem anyhow? Snapshot of the original problem:
Hint. If $\langle n \rangle=\text{round}(\sqrt{n})$ ("round" is also called the nearest integer function) then $$\begin{align}\sum_{n=1}^{\infty} \dfrac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}&=\sum_{k=1}^{\infty}\sum_{j=1}^{2k}\frac{2^k+2^{-k}}{2^{k^2-k+j}} =\sum_{k=1}^{\infty}\frac{2^k+2^{-k}}{2^{k^2-k}}(1-2^{-2k}) \\ &=2\sum_{k=1}^{\infty}(2^{-(k-1)^2}-2^{-(k+1)^2}).\end{align}$$ Now, in order to obtain the final result $3$, note that the last sum is telescopic.