How to prove that $\sum_{k=1}^n\frac{1}{\sqrt[n]{k!} }\sim \frac{n}{\ln n}$ Recently I have met this asymptotic estimation:
prove that $\sum_{k=1}^n\frac{1}{\sqrt[n]{k!} }\sim \frac{n}{\ln n}$，$n\to \infty$.
Here is the way I think:
According to area principle，
\begin{align*}
 \Bigg|\sum_{k=1}^n\frac{1}{\sqrt[n]{k!} }-\int_1^n\frac{1}{\sqrt[n]{\Gamma(x+1)}}d x\Bigg|\leqslant \frac{1}{\sqrt[n]{\Gamma(n+1)}}
\end{align*}
Thus we have
\begin{align*}
\frac{\ln n}{n}\Bigg|\sum_{k=1}^n\frac{1}{\sqrt[n]{k!} }-\int_1^n\frac{1}{\sqrt[n]{\Gamma(x+1)}}d x\Bigg|\leqslant \frac{\ln n}{n}\frac1{\sqrt[n]{\Gamma(n+1)}}\sim \frac{\ln n}{n}\cdot \frac1{\sqrt[n]{2\pi n}} \frac{e}{n}\to 0,~~n\to \infty
\end{align*}
To this end we only need to show
\begin{align*}
\int_1^n\frac{1}{\sqrt[n]{\Gamma(x+1)}}d x\sim \frac{n}{\ln n}
\end{align*}
 But my train of thought is stuck here.Can you give me any help? Thanks a lot.

Using the stirling formula and partial integration we get:
$\displaystyle\frac{\ln n}{n}\sum\limits_{k=1}^n \frac{1}{k!^{1/n}} \sim \frac{\ln n}{n}\int\limits_1^n\frac{dx}{\Gamma(1+x)^{1/n}} = (\ln n)\int\limits_{1/n}^1\frac{dx}{\Gamma(1+nx)^{1/n}}
\sim  (\ln n)\int\limits_{1/n}^1\frac{dx}{(nx/e)^x}$
$\displaystyle = \left(\frac{\ln n}{(x/e)^x}\frac{n^{-x}}{-\ln n}\right)\bigg|_{1/n}^1 + \int\limits_{1/n}^1\frac{\ln \frac{1}{x}}{(nx/e)^x}dx \enspace$ with 
$\enspace\displaystyle \left(\frac{\ln n}{(x/e)^x}\frac{n^{-x}}{-\ln n}\right)\bigg|_{1/n}^1 \sim 1$
$\displaystyle 0< \int\limits_{1/n}^1\frac{\ln \frac{1}{x}}{(nx/e)^x}dx < e\int\limits_0^1\frac{\ln \frac{1}{x}}{n^x}dx =\frac{e}{\ln n}\left(\gamma+\ln\ln n+\int\limits_{\ln n}^\infty\frac{dt}{te^t}\right)\sim\frac{e\ln\ln n}{\ln n}\sim 0$ 
with the Euler-Mascheroni constant $\,\gamma\,$   
It follows $\enspace\displaystyle\frac{\ln n}{n}\sum\limits_{k=1}^n \frac{1}{k!^{1/n}} \sim 1\enspace$ and therefore the claim.
Note:
$\displaystyle \int\limits_0^1\frac{\ln \frac{1}{x}}{e^{ax}}dx=\frac{\ln a + \gamma + \Gamma(0,a)}{a}\sim \frac{\ln a}{a}\enspace$ with the incomplete Gamma function $\,\Gamma(.,.)$
