How to find the limit:$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$ How to find the limit:$$\lim_{n\to \infty}\left(\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\right)$$
I can't think of any way of this problem
Can someone to evaluate this?
Thank you.

For small $x>0$, we have $1+x\leq\exp(x)\leq 1+x+x^{2}$, then for large $k$,
\begin{align*}
1+\dfrac{\log 2}{k}-\left(\dfrac{\log k}{k}\right)^{2}\leq 2(2k)^{1/2k}-k^{1/k}\leq 1+\dfrac{\log 2}{k}+\dfrac{(\log 2k)^{2}}{2k^{2}},
\end{align*}
and for large $n$,
\begin{align*}
\left(\sum_{k=n+1}^{2n}2(2k)^{1/2k}-k^{1/k}\right)-n\leq\log 2\sum_{k=n+1}^{2n}\dfrac{1}{k}+\sum_{k=n+1}^{2n}\dfrac{(\log 2k)^{2}}{2k^{2}},
\end{align*}
and
\begin{align*}
\left(\sum_{k=n+1}^{2n}2(2k)^{1/2k}-k^{1/k}\right)-n\geq\log 2\sum_{k=n+1}^{2n}\dfrac{1}{k}-\sum_{k=n+1}^{2n}\left(\dfrac{\log k}{k}\right)^{2},
\end{align*}
and note that 
\begin{align*}
\sum_{k=1}^{\infty}\dfrac{(\log 2k)^{2}}{2k^{2}}&<\infty,\\
\sum_{k=1}^{\infty}\left(\dfrac{\log k}{k}\right)^{2}&<\infty,
\end{align*}
and we treat 
\begin{align*}
\sum_{k=n+1}^{2n}\dfrac{1}{k}
\end{align*}
as
\begin{align*}
\int_{n+1}^{2n}\dfrac{1}{t}dt=\log 2n-\log(n+1)=\log\left(\dfrac{2n}{n+1}\right)
\end{align*}
when $n\rightarrow\infty$. So the limit is $(\log 2)^{2}$. 
