Limit when $n$ tends to $\infty$ of $\log(n!)/\log(n+1)$?

Question

I've been trying for several days now to find the limit of $\dfrac{\log(n!)}{\log(n+1)}$ when $n$ tends to $\infty$.

I have no idea how to continue... I've tried to ask in other forums but I always get answers that only hint to the solution. I would really appreciate to get an explicit solution if possible since I'm stuck...

(The main goal is to find the radius of convergence of the complex power series where $a_{n}=\log(n!)$)

Thank you!

Answer 1

[Just to offer a more elementary answer than Jack D'Auricio's comments:]

Let $k$ be a fixed natural number. The estimation $$\frac{\log n!}{\log(n+1)}\geq\sum_{j=1}^k\frac{\log(n+1-j)}{\log(n+1)},$$ for $n\geq k$, gives us that $$\lim_{n\to\infty}\frac{\log n!}{\log(n+1)}\geq k.$$ As $k$ was arbitrary it follows that $$\lim_{n\to\infty}\frac{\log n!}{\log(n+1)}=+\infty. $$

Answer 2

With some log properties and Riemann sums:

$$\log(n!)=\sum_{k=1}^n\log(k)>\int_1^n\log(x)\ dx=n\log(n)-n+1$$

Thus,

$$\frac{\log(n!)}{\log(n+1)}>\frac{n\log(n)-n+1}{\log(n+1)}$$

whereupon we find divergence as $n\to\infty$ quite easily with L'Hospital's rule, some basic expansion, etc.

Answer 3

For $k\ge2$ we have the following:

$$1-{1\over k}\le{1\over2}+{1\over3}+\cdots+{1\over k}\le\int_1^k{dx\over x}=\ln k$$

(The first inequality is easily checked for $k=2$ and $3$, and holds for $k\ge4$ since ${1\over2}+{1\over3}+{1\over4}\gt1$.) Thus

$$\begin{align} \ln(n!) &=\ln2+\ln3+\cdots+\ln n\\ &\ge\left(1-{1\over2}\right)+\left(1-{1\over3}\right)+\cdots+\left(1-{1\over n}\right)\\ &=n-1-\left({1\over2}+{1\over3}+\cdots+{1\over n}\right)\\ &\ge n-1-\ln n \end{align}$$

so that

$${\ln(n!)\over\ln(n+1)}\ge{n-1\over\ln(n+1)}-1\to\infty\quad\text{as }n\to\infty$$

Answer 4

Since $n! > (n/e)^n$ (proved by induction from $(1+1/n)^n < e < (1+1/n)^{n+1}$), $\log n! > n(\log n - 1)$ so, an even stronger result holds: $\frac{\log(n!)}{n} \gt \log n - 1 \to \infty $.