Limit of $\log(n!)/\log((n+1)!)$? I want to show that $$\lim_{n\to\infty}\frac{\log(n!)}{\log((n+1)!)}=1.$$
Obviously, $$\frac{\log(n!)}{\log((n+1)!)}=\frac{\log(n!)}{\log(n+1)+\log(n!)}\leq1.$$
But I get stuck with the other estimate. It looks very similar to other problems prior to this, but I cannot derive a converging lower bound. I am open to fundamental proofs and proofs using Stirling's formula.
 A: What you try to prove is equivalent to:
$\lim_{n\to\infty}\frac{\log((n+1)!)}{\log(n!)}=1.$
Substracting $1$ at both sides, that´s the same as:
$\lim_{n\to\infty}\frac{\log(n+1)}{\log(n!)}=0$.
But for a given $k>0$, there is some $N$ such that for any $n>N$, you have $(n+1)^k<n!$. For example, you can pick $N=3k$, and then for $n>N$, $(n+1)^k<(2(n-k))^k<2^k(n-k)^k<2\cdot3\dots\cdot k\cdot (n-k)\cdot\dots\cdot (n-1)n<n!$.
So for every $k$ there is $N$ such that for each $n>N$, $\frac{\log(n+1)}{\log(n!)}<\frac{1}{k}$. Thus, $\frac{\log(n+1)}{\log(n!)}$ converges to $0$, as we were trying to prove.
A: How about $$\displaystyle\lim_{n \to \infty} \frac{\log(n!)}{\log((n + 1)!)} = \lim_{n \to \infty} \frac{\log(n!)}{\log(n + 1) + \log(n!)} = \lim_{n \to \infty} \left( \displaystyle 1 - \frac{\log(n + 1)}{\log(n + 1) + \log(n!)} \right) = 1$$
A: Hint $\log(n!)=\log(\frac{(n+1)!}{n+1})=\log((n+1)!)-\log(n+1)$
A: HINT add and subtract $\log(n+1)$ to the numerator, to get something like:
$$\frac{\log(n!)+\log(n+1)-\log(n+1)}{\log((n+1)!)}=1-\frac{\log(n+1)}{\log((n+1)!)}$$
