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.

Solution 1:

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 $\,\Gamma(.,.)$

Solution 2:

For any $x>0$, $$c_1\,x^{x+\frac{1}{2}}e^{-x}\leq \Gamma(x+1)\leq c_2\,x^{x+\frac{1}{2}}e^{-x}$$ for some $c_1,c_2>0$ (see for example the Wikipedia page). Therefore, $$\frac{1}{\sqrt[n]{c_2}}\int_1^nx^{-x/n}x^{-\frac{1}{2n}}e^{x/n}\,dx\leq \int_1^n\frac{1}{\sqrt[n]{\Gamma(x+1)}}\,dx\leq\frac{1}{\sqrt[n]{c_1}}\int_1^nx^{-x/n}x^{-\frac{1}{2n}}e^{x/n}\,dx.$$ Next, note that $$n^{-\frac{1}{2n}}\int_1^nx^{-x/n}e^{x/n}\,dx\leq\int_1^nx^{-x/n}x^{-\frac{1}{2n}}e^{x/n}\,dx\leq \int_1^nx^{-x/n}e^{x/n}\,dx,$$ so it suffices to find the asymptotics of the last integral.

Solution 3:

The terms $\frac{1}{\sqrt[n]{k!}}$, for $k\in[1,n]$, are roughly of the same size, hence the upper bound provided by Holder's inequality $$ \sum_{k=1}^{n}\frac{1}{\sqrt[n]{k!}}\leq \sqrt[n]{n^{n-1}\sum_{k=1}^{n}\frac{1}{k!}}\leq \frac{n}{\sqrt[n]{n/e}}$$ is expected to be close to the asymptotic behaviour of the RHS.
It can be improved via (I am going to outline the case $n=4$ for simplicity) $$\sum_{k=1}^{4}\frac{1}{\sqrt[4]{k!}}\leq \sqrt[4]{\left(\tfrac{1}{1}+\tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}\right)\left(\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{2}+\tfrac{1}{3}\right)\left(\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{2}\right)\left(\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{1}+\tfrac{1}{1}\right)}$$ leading to $$ \sum_{k=1}^{n}\frac{1}{\sqrt[n]{k!}}\leq \sqrt[n]{\prod_{h=0}^{n-1}\left(h+H_{n-h}\right)}.\tag{U}$$ On the other hand Holder's inequality can be used also for producing a lower bound:

$$ \left(\sum_{k=1}^{n}\frac{1}{1}\right)\left(\sum_{k=1}^{n}\frac{1}{\sqrt[n]{k!}}\right)^n \geq \left(\sum_{k=1}^{n}\frac{1}{\sqrt[n+1]{k!}}\right)^{n+1}$$ $$ \left(\sum_{k=1}^{n}\frac{1}{1}\right)\left(\sum_{k=1}^{n}\frac{1}{\sqrt[n+1]{k!}}\right)^{n+1} \geq \left(\sum_{k=1}^{n}\frac{1}{\sqrt[n+2]{k!}}\right)^{n+2}$$

lead to $$ \sum_{k=1}^{n}\frac{1}{\sqrt[n]{k!}}\geq \frac{1}{n^{1/n}}\left(\sum_{k=1}^{n}\frac{1}{\sqrt[n+1]{k!}}\right)^{\frac{n+1}{n}}\geq \frac{1}{n^{2/n}}\left(\sum_{k=1}^{n}\frac{1}{\sqrt[n+2]{k!}}\right)^{\frac{n+2}{n}}\geq\ldots\geq\frac{1}{n}\left(\sum_{k=1}^{n}\frac{1}{\sqrt[2n]{k!}}\right)^2,$$

$$ \sum_{k=1}^{n}\frac{1}{\sqrt[n]{k!}}\geq\frac{1}{n^n}\left(\sum_{k=1}^{n}\frac{1}{\sqrt[n^2]{k!}}\right)^{n+1}\tag{L}$$ and for any $k\in[1,n]$ the distance between $\sqrt[2n]{k!}$ and $1$ is $O\left(\frac{\log n}{n}\right)$ by Stirling's approximation. The asymptotic behaviour of your sum should now follow by comparing the accurate upper bound $(U)$ and the accurate lower bound $(L)$.