Sunday, 27 January 2019

calculus - Finding the limit of fracnsqrt[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.


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