Vtyjkyuk

Once I met a problem about limits: $$lim_nrightarrowinftyint_0^infty e^-xfracln^n frac1xn! dx=1$$

After I proved it, I used Mathematica and discovered that the sequence $int_0^infty e^-xfracln^nfrac1xn! dx$ seems to be monotone increasing to its supremum $1$.

I guess so, but I’m confused about how to prove.

My question is: How to prove that $$int_0^infty e^-xln^n frac1xdx<n!,quad forall nge 1$$

and the monotonity of $$int_0^infty e^-xfracln^nfrac1xn! dx$$

Additional: I have already proved that $$lim_nrightarrowinfty frac2^n+1n!left(n!-int_0^infty e^-xln^n frac1xdxright)=lim_nrightarrowinfty frac2^n+1n!left(int_0^1 left(1-e^-xright)ln^nfrac1x dx-int_1^infty e^-xln^nfrac1x dxright)=1$$

by using the inequality $x-fracx^22le 1-e^-xle x$ and $ln^n x<n! xleft(xge 1right)$.

It can indicate that $1-int_0^infty e^-xfracln^n frac1xn! dx$ is EVENTUALLY (which means for sufficient large $n$) positive and monotone decreasing at $Oleft(2^-nright)$.

I think you can go with this walk:

Let be $f(x) = e^-x ln^nfrac1x$ consider the function $g(x) = e^-xx^n$.

We know that $ int_0^infty g(x) dx = int_0^inftye^-xx^n dx = Gamma(n+1) = n!$

$f(x) < g(x) Rightarrow int_0^infty f(x) dx < int_0^infty g(x) dx = n!$

for proving this, we need to prove that $ln^n frac1x< x^n$. So, let be $h(x) = x^n - ln ^n frac1x$

$h'(x)> 0 Rightarrow n x^n-1 - (n ln^n-1(frac1x))frac1frac1x(-frac1x^2) = n(x^n-1 + frac1xln^n-1(frac1x)) > 0 Rightarrow x^n > (-1)^n-1 ln ^n-1x$

I think it done.