Evaluate the series $\sum_{n=1}^{\infty} \frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}$

How can I obtain the limit of the series

$$\sum_{n=1}^{\infty} \frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}$$

Where $[\ \ ]$ is Nearest Integer Function.

Solution 1:

For $(k-0.5)^2<k^2-k+1\le n\le k^2+k<(k+0.5)^2$ we have $[\sqrt{n}]=k$. Furthermore, $(k+1)^2-(k+1)+1=k^2+k+1$ and since $1^2-1+1=1$ we have $\bigcup_{k=1}^{\infty}\left[k^2-k+1;…;k^2+k\right]=\mathbb{N}$. Therefore, since the sum clearly converges, we have: $$ \sum_{n=1}^{\infty} \frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}=\sum_{k=1}^{\infty}\left[\sum_{n=k^2-k+1}^{k^2+k} \frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}\right]=\sum_{k=1}^{\infty}\left[\sum_{n=k^2-k+1}^{k^2+k} \frac{2^{k}+2^{-k}}{2^n}\right]=\sum_{k=1}^{\infty}\left[(2^{k}+2^{-k})\sum_{n=k^2-k+1}^{k^2+k} \frac{1}{2^n}\right]=\sum_{k=1}^{\infty}\left[\left(2^{k}+2^{-k}\right)\left(2\cdot\left(1-2^{-(k^2+k+1)}\right)-2\cdot\left(1-2^{-(k^2-k+1)}\right)\right)\right]=2\cdot\sum_{k=1}^{\infty}\left[\left(2^{k}+2^{-k}\right)\left(-2^{-(k^2+k+1)}+2^{-(k^2-k+1)}\right)\right]=2\cdot\sum_{k=1}^{\infty}\left[2^{-(k-1)^2}-2^{-(k+1)^2}\right] $$ and the result follows by telescoping.

Solution 2:

This is not a complete answer, but perhaps it can help and/or motivate others to find and prove the answer.

I did a numerical computation using $1000$ terms and I got a result of $3$ with very high precision. Hence I conjecture that

$$\sum_{n=1}^\infty\frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}=3$$