# How do I find a relation between the factorial function and $n^{\text{th}}$ root while solving for limits?

The title of this question mostly says it all. I'm learning sequences and while finding limits, I specifically struggle when I have some combination of $$\sqrt[n]{\cdot}$$ and $$n!$$, I seem to struggle because I don't know how to relate the two things.

For example, take the simplest combination $$a_n= \sqrt[n]{n!}$$. Now what do I do with this thing?

$$\lim\limits_{n\to ∞} \sqrt[n]{n!} = \lim\limits_{n\to ∞} \exp\left(\dfrac{\sum\limits_{k=1}^{n}\ln(k)}{n}\right)$$

Marvelous, now I have another nightmare! But I know since $$\ln(n)\to ∞$$ then it's sequence of average $$\frac{\sum\limits_{k=1}^{n}\ln(k)}{n}$$ also diverges to $$∞$$, so I'm saved. Therefore $$a_n \to ∞$$

But what do I do when, say I have something like,

$$\lim\limits_{n\to∞} \dfrac{\ln(n+1)!}{\ln(n!)}$$ or $$\lim\limits_{n\to∞} \dfrac{\ln((n+1)!)}{(n!)^{1/n}}$$

... or some combination of those things. You get the idea. So how you do suggest I proceed in these cases? If you can share the general ideas on how to go about solving those guys or just demonstrate maybe, it'd be very helpful.

One extremely helpful substitution that is often ignored at an elementary level is Stirling's approximation, which says: $$n!\sim \sqrt{2\pi n} {\left(\frac ne\right)}^n$$ Note how easy it becomes to use this to calculate the popular limit: $$L=\lim_{n\to \infty} \frac {n}{(n!)^{\frac 1n}}$$ We get: $$L=\lim_{n\to \infty} \frac {e}{(2\pi n)^{\frac {1}{2n}}}$$ Which means that $$L=e$$, since $$\lim_{n\to \infty} n^{\frac 1n}=1$$ The approximation is also highly accurate for even moderately large values of $$n$$.
• This is mad awesome! I'll read more about it. Are there any more of those kinds of approximations for large $n$, you can share? Jun 14 at 17:25
• We have:$$\lim_{n\to \infty} \frac 1n \sum_{r=1}^n f(\frac rn)=\int_{0}^1 f(x) dx$$ This fact could be used, since taking antilog on $n!$ converts it into a sum of $\ln$. Jun 14 at 19:21
• While sniffing around a little, I found out a more general version of that, though I don't know how to prove it yet but, $$\lim\limits_{n \to ∞} \dfrac{1}{n} \sum\limits_{r=a}^{kn} f \left(\dfrac{r}{n}\right) = \int_0^{k} f(x) \, \mathrm{d}x, a,k \in \mathbb{Z}$$ Jun 15 at 21:14