If $$f(n)=\left[\sqrt{n}+\frac{1}{2}\right]$$ where $[\cdot]$ denotes the greatest integer function, then find the value of the limit $$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{2^{f(k)}+2^{-f(k)}}{2^{n}}\right).$$
My attempt: I tried to evaluate individual values of $f(n)$ for different values of n to get a pattern, e.g. $$f(1) = f(2)= 1,\quad f(3) = f(4)= f(5) = f(6)= 2,\\ f(7) = f(8) = f(9) = f(10) = f(11) = f(12) = 3,\dots $$ but I feel there should be a generic way of handling this problem.
Any help will be appreciated.
We have that $$0\leq \frac{1}{2^{n}}\sum_{k=1}^n\left(2^{f(k)}+2^{-f(k)}\right)\leq \frac{n\left(2^{f(n)}+1\right)}{2^{n}}\leq \frac{n\left(2^{\sqrt{n}+\frac{1}{2}}+1\right)}{2^{n}}\to 0.$$ Therefore your limit is zero: $$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{2^{f(k)}+2^{-f(k)}}{2^{n}}\right)=0.$$
Maybe you are interested in this other limit: $$L_2:=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{2^{f(k)}+2^{-f(k)}}{2^{k}}\right)$$ Then note that $\{f(n)\}_{n\geq 1}$ is an increasing sequence where the number $k$ appears $2k$ times.
Hence $$L_2=\sum_{k=1}^{\infty}\left(2^{k}+2^{-k}\right)\sum_{j=k(k-1)+1}^{k(k+1)}\frac{1}{2^j}=\sum_{k=1}^{\infty}\sum_{i=(k-1)^2}^{k^2}\frac{1}{2^i} +\sum_{k=1}^{\infty}\sum_{i=k^2+1}^{(k+1)^2-1}\frac{1}{2^i}\\ =\sum_{k=1}^{\infty}\sum_{i=(k-1)^2}^{k^2-1}\frac{1}{2^i} +\sum_{k=1}^{\infty}\sum_{i=k^2}^{(k+1)^2-1}\frac{1}{2^i} =\sum_{i=0}^{\infty}\frac{1}{2^i} +\sum_{i=1}^{\infty}\frac{1}{2^i}=2+1=3.$$