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.

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 $0$.
