Vtyjkyuk

Recently I have met this asymptotic estimation:
prove that $sum_k=1^nfrac1sqrt[n]k! sim fracnln n$，$nto infty$.

Here is the way I think:

According to area principle，
beginalign*
Bigg|sum_k=1^nfrac1sqrt[n]k! -int_1^nfrac1sqrt[n]Gamma(x+1)d xBigg|leqslant frac1sqrt[n]Gamma(n+1)
endalign*
Thus we have
beginalign*
fracln nnBigg|sum_k=1^nfrac1sqrt[n]k! -int_1^nfrac1sqrt[n]Gamma(x+1)d xBigg|leqslant fracln nnfrac1sqrt[n]Gamma(n+1)sim fracln nncdot frac1sqrt[n]2pi n fracento 0,~~nto infty
endalign*
To this end we only need to show
beginalign*
int_1^nfrac1sqrt[n]Gamma(x+1)d xsim fracnln n
endalign*
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:

$displaystylefracln nnsumlimits_k=1^n frac1k!^1/n sim fracln nnintlimits_1^nfracdxGamma(1+x)^1/n = (ln n)intlimits_1/n^1fracdxGamma(1+nx)^1/n
sim (ln n)intlimits_1/n^1fracdx(nx/e)^x$

$displaystyle = left(fracln n(x/e)^xfracn^-x-ln nright)bigg|_1/n^1 + intlimits_1/n^1fracln frac1x(nx/e)^xdx enspace$ with
$enspacedisplaystyle left(fracln n(x/e)^xfracn^-x-ln nright)bigg|_1/n^1 sim 1$

$displaystyle 0< intlimits_1/n^1fracln frac1x(nx/e)^xdx < eintlimits_0^1fracln frac1xn^xdx =fraceln nleft(gamma+lnln n+intlimits_ln n^inftyfracdtte^tright)simfracelnln nln nsim 0$

with the Euler-Mascheroni constant $,gamma,$

It follows $enspacedisplaystylefracln nnsumlimits_k=1^n frac1k!^1/n sim 1enspace$ and therefore the claim.

Note:

$displaystyle intlimits_0^1fracln frac1xe^axdx=fracln a + gamma + Gamma(0,a)asim fracln aaenspace$ with the incomplete Gamma function $,Gamma(.,.)$

For any $x>0$, $$c_1,x^x+frac12e^-xleq Gamma(x+1)leq c_2,x^x+frac12e^-x$$ for some $c_1,c_2>0$ (see for example the Wikipedia page). Therefore, $$frac1sqrt[n]c_2int_1^nx^-x/nx^-frac12ne^x/n,dxleq int_1^nfrac1sqrt[n]Gamma(x+1),dxleqfrac1sqrt[n]c_1int_1^nx^-x/nx^-frac12ne^x/n,dx.$$ Next, note that $$n^-frac12nint_1^nx^-x/ne^x/n,dxleqint_1^nx^-x/nx^-frac12ne^x/n,dxleq int_1^nx^-x/ne^x/n,dx,$$ so it suffices to find the asymptotics of the last integral.

The terms $frac1sqrt[n]k!$, for $kin[1,n]$, are roughly of the same size, hence the upper bound provided by Holder's inequality
$$ sum_k=1^nfrac1sqrt[n]k!leq sqrt[n]n^n-1sum_k=1^nfrac1k!leq fracnsqrt[n]n/e$$
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)
$$sum_k=1^4frac1sqrt[4]k!leq sqrt[4]left(tfrac11+tfrac12+tfrac13+tfrac14right)left(tfrac11+tfrac11+tfrac12+tfrac13right)left(tfrac11+tfrac11+tfrac11+tfrac12right)left(tfrac11+tfrac11+tfrac11+tfrac11right)$$
leading to
$$ sum_k=1^nfrac1sqrt[n]k!leq sqrt[n]prod_h=0^n-1left(h+H_n-hright).tagU$$
On the other hand Holder's inequality can be used also for producing a lower bound:

$$ left(sum_k=1^nfrac11right)left(sum_k=1^nfrac1sqrt[n]k!right)^n geq left(sum_k=1^nfrac1sqrt[n+1]k!right)^n+1$$
$$ left(sum_k=1^nfrac11right)left(sum_k=1^nfrac1sqrt[n+1]k!right)^n+1 geq left(sum_k=1^nfrac1sqrt[n+2]k!right)^n+2$$

lead to
$$ sum_k=1^nfrac1sqrt[n]k!geq frac1n^1/nleft(sum_k=1^nfrac1sqrt[n+1]k!right)^fracn+1ngeq frac1n^2/nleft(sum_k=1^nfrac1sqrt[n+2]k!right)^fracn+2ngeqldotsgeqfrac1nleft(sum_k=1^nfrac1sqrt[2n]k!right)^2,$$

$$ sum_k=1^nfrac1sqrt[n]k!geqfrac1n^nleft(sum_k=1^nfrac1sqrt[n^2]k!right)^n+1tagL$$
and for any $kin[1,n]$ the distance between $sqrt[2n]k!$ and $1$ is $Oleft(fraclog nnright)$ 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 $(L)$.