I'm trying to find
limn→∞nn√n!.
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=limn→∞n√an=limn→∞an+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 infn→∞cn+1cn≤lim infn→∞n√cn,
lim supn→∞n√cn≤lim supn→∞cn+1cn.
In the present context, this shows that lim infn→∞(1+1n)n≤lim infn→∞nn√n!≤lim supn→∞nn√n!≤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 limn→∞cn+1cn exists follows from the Stolz–Cesàro theorem applied to finding the limit of ln(cn)n. Explicitly,
limn→∞ln(n√cn)=limn→∞ln(cn)n=limn→∞ln(cn+1)−ln(cn)(n+1)−n=limn→∞ln(cn+1cn),
provided the latter limit exists, where the second equality is by the Stolz–Cesàro theorem.
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