12

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.

Caleb Stanford
  • 45,895
Afonso Henriques
  • 223
  • 3
    Please don't close. This is interesting. – Spenser Dec 30 '14 at 15:49
  • 2
    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? –  Dec 30 '14 at 21:09
  • 3
    I considered this interesting enough to ask again. http://math.stackexchange.com/questions/1086984/a-curious-summation-that-appears-to-be-3 – MJD Dec 31 '14 at 19:55

2 Answers2

13

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.

Redundant Aunt
  • 12,030
  • 2
  • 22
  • 66
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

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

Spenser
  • 19,469