Finding the limit of nn√n!\frac {n}{\sqrt[n]{n!}}

I’m trying to find
limnnnn!.

I tried couple of methods: Stolz, Squeeze, D’Alambert

Thanks!

Edit: I can’t use Stirling.

Answer

Let an=nnn!. Then the power series n=1anxn has radius of convergence R satisfying 1R=limnnan=limnan+1an, provided these limits exist. The first limit is what you’re looking for, and the second limit is limn(1+1n)n.

Added: I just happened upon a good reference for the equality of limits above, which gives a more general result which is proved directly without reference to power series. Theorem 3.37 of Rudin’s Principles of mathematical analysis, 3rd Ed., says:

For any sequence {cn} of positive numbers,
lim infncn+1cnlim infnncn,
lim supnncnlim supncn+1cn.

In the present context, this shows that lim infn(1+1n)nlim infnnnn!lim supnnnn!lim supn(1+1n)n.
Assuming you know what limn(1+1n)n is, this shows both that the limit in question exists (in case you didn’t already know by other means) and what it is.


From the comments: User9176 has pointed out that the case of the theorem above where limncn+1cn exists follows from the Stolz–Cesàro theorem applied to finding the limit of ln(cn)n. Explicitly,
limnln(ncn)=limnln(cn)n=limnln(cn+1)ln(cn)(n+1)n=limnln(cn+1cn),
provided the latter limit exists, where the second equality is by the Stolz–Cesàro theorem.

Attribution
Source : Link , Question Author : Community , Answer Author : Jonas Meyer

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