How to Compute:$\lim\limits_{n\to+\infty}n(1+n+ \ln{n}-\sum\limits_{k=1}^{n}\sqrt[n]{k})$ I need Help to Compute:
$$\lim_{n\to+\infty}n(1+n+ \ln{n}-\sum_{k=1}^{n}\sqrt[n]{k})$$+
I have tired to squeeze the general but so far I got
that $$\sum_{k=1}^n \sqrt[n]{k}\geq \frac{n^{\frac{2n+1}{n}}}{n+1}$$
But I don't how to continue from here. Any idea?
 A: The obvious upper bound is $$\sum_{k=1}^n \sqrt[n]{k} \le n^{1 + 1/n}$$
A: $$\sqrt[n]{k}=\exp\left(\frac{\log k}{n}\right)=1+\frac{\log k}{n}+\frac{\log^2 k}{2n^2}+O\left(\frac{\log^3 k}{n^3}\right) \tag{1}$$
$$ \sum_{k=1}^{n}\sqrt[n]{k} = n+\frac{1}{n}\log(n!)+\frac{1}{2n^2}\sum_{k=1}^{n}\log^2(k)+O\left(\frac{\log^3 n}{n^3}\right)\tag{2}$$
$$\large\scriptstyle \sum_{k=1}^{n}\sqrt[n]{k} = n+\left(1\color{red}{+\frac{1}{2n}}\right)\log(n)\color{red}{-1}+\frac{\log\sqrt{2\pi}}{n}+\frac{1}{12 n^2}+\frac{1}{2n^2}\sum_{k=1}^{n}\log^2(k)+O\left(\frac{\log^3 n}{n^3}\right)\tag{3}$$
By Stirling's approximation. Due to the presence of the highlighted terms, the wanted limit equals $+\infty$.
A: The Euler-Maclaurin Sum Formula, we get
$$
\begin{align}
&n\sum_{k=1}^n\left(k^{1/n}-1\right)\\
&=n\sum_{k=1}^n\left(e^{\log(k)/n}-1\right)\\
&=\sum_{j=1}^\infty\frac1{j!n^{j-1}}\sum_{k=1}^n\log(k)^j\\
&=\left(n\log(n)-n+\frac12\log(n)\right)+\left(\frac12\log(n)^2-\log(n)+1\right)+O\!\left(\frac{\log(n)^3}n\right)\\
&=n\log(n)-n+\frac12\log(n)^2-\frac12\log(n)+1+O\!\left(\frac{\log(n)^3}n\right)
\end{align}
$$
Therefore,
$$
\begin{align}
n\left(1+n+\log(n)-\sum_{k=1}^n\sqrt[n]{k}\right)
&=n+n\log(n)-n\sum_{k=1}^n\left(k^{1/n}-1\right)\\
&=2n-\frac12\log(n)^2+\frac12\log(n)-1+O\!\left(\frac{\log(n)^3}n\right)
\end{align}
$$
Thus, the term grows like $2n$, so the limit is $\infty$.
