Limit of logarithm function, $\lim\limits_{n\to\infty}\frac{n}{\log \left((n+1)!\right)}$ Determine $$\lim_{n \to \infty} \frac{n}{\log \left((n+1)!\right)}.$$
Now, I know that $\log x < \sqrt{x} < x$ and trying to apply comparison test, but it does not work. Please help.
 A: Note that
$$(n!)^2=\Bigl(1\cdot n\Bigr)\Bigl(2\cdot(n-1)\Bigr)\Bigl(3\cdot(n-2)\Bigr)\ldots\Bigl((n-1)\cdot 2\Bigr)\Bigl(n\cdot 1\Bigr)\ge n\cdot n\cdot\ldots\cdot n=n^n$$
(the graph of $x\cdot((n+1)-x)$ is a parabola oriented downwards)
Therefore (for $n\ge 2$)
$$0\le\frac{n}{\log((n+1)!)}\le\frac{n}{\log(n!)}=\frac{2n}{\log((n!)^2)}\le\frac{2n}{\log{n^n}}=\frac{2}{\log n}\to0$$
and the limit is $0$ by the squeeze theorem.
A: We have that
$$
\lim_{n\to\infty}\frac n{\log(n+1)!}=\lim_{n\to\infty}\frac n{\log 2+\log3+\ldots+\log(n+1)}.
$$
Since the logarithm is a monotone function, we can estimate the sum with integrals from below and above in the following way
$$
\int_1^{n+1}\log x\mathrm dx\le\sum_{i=2}^{n+1}\log i\le\int_2^{n+2}\log x\mathrm dx.
$$
Also, we have that
$$
\int_1^{n+1}\log x\mathrm dx=(n+1)(\log(n+1)-1)+1
$$
and
$$
\int_2^{n+2}\log x\mathrm dx=(n+2)(\log(n+2)-1)-2(\log 2-1).
$$
Hence,
$$
\sum_{i=2}^{n+1}\log i\sim n\log n
$$
as $n\to\infty$, where $\sim$ means that the ratio of the sequences tend to $1$ as $n\to\infty$. So the limit in question is $0$.
A: By Stirling's approximation $\log n!=O(n\log n)$.
Also $O((n+1)\log(n+1))=O(n\log n)$.
Therefore, $$\lim_{n\to\infty}\frac{n}{\log((n+1)!)}=\lim_{n\to\infty}\frac{n}{n\log n}=\lim_{n\to\infty}\frac{1}{\log n}=0$$.
