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.

  • 3
    $\begingroup$ Please don't close. This is interesting. $\endgroup$
    – Spenser
    Dec 30, 2014 at 15:49
  • 2
    $\begingroup$ Dear Afonso: please try read the article How to ask a Good Question?), especially the part that emphasizes context. Where is this problem from? What makes it difficult for you to solve? Since you know that the nearest integer function does, presumably you knew that it's constant within certain ranges? $\endgroup$
    – user147263
    Dec 30, 2014 at 21:09
  • 3
    $\begingroup$ I considered this interesting enough to ask again. math.stackexchange.com/questions/1086984/… $\endgroup$
    – MJD
    Dec 31, 2014 at 19:55

2 Answers 2


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.


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


  • $\begingroup$ I ran a simulation as well and reached the same result. $\endgroup$
    – rubik
    Dec 30, 2014 at 16:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .