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.

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