# 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 $(\log 2)^{2}$.
$$k^{\frac1k}=e^{\frac{\log k}{k}}\sim1+\frac{\log k}{k}$$
$$2k^{\frac{1}{2k}}=e^{\frac{\log 2k}{2k}}\sim1+\frac{\log 2k}{2k}$$
$$2k^{\frac{1}{2k}}-k^{\frac1k}\sim1+\frac{\log 2}{k}$$
$$\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)\sim n+\log 2\sum_{k=n+1}^{2n}\frac1k\sim n+\log 2 \, \log \left(\frac{2n}{n+1}\right)$$
$$\sum_{k=n+1}^{2n}\left(2(2k)^{\frac{1}{2k}}-k^{\frac{1}{k}}\right)-n\sim \log 2 \, \log \left(\frac{2n}{n+1}\right)\to(\log2)^2$$