Evaluate $\underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}\right)$

Question

Evaluate the limit $$ \underset{n\to \infty }{\text{lim}}\left(\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}\right) $$

I got stuked when trying to evaluate it and I even don't know where to begin. Mathematica tells me that $$ \sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n}=\frac{2^{1/n} \left(-\left(2^{1/n}\right)^n \Phi \left(2^{1/n},1,n+1+\frac{1}{n}\right)+\left(2^{1/n}\right)^{n+1} \Phi \left(2^{1/n},1,n+1+\frac{1}{n}\right)-2^{1/n} \Phi \left(2^{1/n},1,1+\frac{1}{n}\right)+\Phi \left(2^{1/n},1,1+\frac{1}{n}\right)+n \left(2^{1/n}\right)^n-n\right)}{\left(2^{1/n}-1\right) n^2} $$ However it doesn't give the result of the limit either. And I tried $n=10000$ numerically, the answer is $1.44274..$.

So how to calculate the limit?

Answer 1

We have $$ \sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{k}+n} = \frac1n\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{nk}+1}$$ and now $$\frac1n\sum _{k=1}^n \frac{2^{k/n}}{\frac{1}{nk}+1}=\frac1n\sum _{k=1}^n 2^{k/n}-\frac1n\sum _{k=1}^n \frac{2^{k/n}}{nk+1}$$ where the first term converges to $\int_0^1 2 ^xdx$ and the second term vanishes.

Answer 2

Your result $R(n)$ in the last line can be approximated in Mathematica using the function

Series[R[n],Infinity,1]

I got $$\frac{1}{2 n}+\frac{1}{\log 2}\to \frac{1}{\log 2}$$