Showing $\lim_{n \to +\infty} \log(n!)/(n\log n) = 1$ without using Stirling approximation As a passage of a bigger limit I have to show that $$ \lim_{ n \to \infty } \frac{\log(n!)}{n\log(n)} = 1. $$
I think it could be done using Stirling approximation, but I'm wondering if there's a way without that formula.
 A: With an integral test for convergence: $\displaystyle \int_1^n \log(x)dx \leq \sum\limits_{k=2}^n \log(k) = \log(n!) \leq \int_2^{n+1} \log(x)dx$.
You can deduce the stronger result: $\log(n!)=n\log(n)-n + o(n)$. 
A: We have
$$\log n!=\sum_{j=1}^n\log\frac jn+n\log n,$$
so we have to show that $\frac 1{n\log n}\sum_{j=1}^n\log\frac jn\to 0.$ We use the inequality $\log(1+t)\geq t-t^2/2$ for $t\geq -1$, to get 
$$0\geq \frac 1{n\log n}\sum_{j=1}^n\log\frac jn\geq \frac 1{n\log n}\sum_{j=1}^n\left(\frac jn-1\right)-\frac 1{2n\log n}\sum_{j=1}^n\left(\frac{j}{n}-1\right)^2.$$
We have in the RHS $\frac 1{\log n}$ times a Riemann sum (associated with $t\mapsto t-1-\left(t-1\right)^2/2$), hence $\frac 1{\log n}$ times a bounded term, which gives the result.
A: This can be done with very little technology. Note that
$$\log n! = \sum_{i=1}^n \log i \le \sum_{i=1}^n \log n = n \log n$$
so it suffices to show that $\log n!$ is eventually greater than $cn \log n$ for every constant $c < 1$. But for every positive constant $k$ we have
$$\log n! = \sum_{i=1}^n \log i \ge \sum_{i=\lfloor \frac{n}{k} \rfloor}^n \log \left\lfloor \frac{n}{k} \right\rfloor \approx \left( 1 - \frac{1}{k} \right) n \left( \log n - \log k \right)$$
and taking $k \to \infty$ the conclusion follows. 
A: We compute $$ \lim_{n\to\infty} \frac{ \log( (n+1)! ) - \log (n!) }{ (n+1)\log(n+1) - n\log n} = \lim_{n\to\infty} \frac{ \log (n+1) }{ \log (n+ \theta_n)+1}=1$$
where $\theta_n \in (0,1)$ is obtained from the Mean Value Theorem. It follows from the Stolz-Cesàro theorem that the limit in question is also equal to $1.$
A: Using
$$\left( \frac{n}{e}\right)^n \lt n! \lt e \left( \frac{n}{2}\right)^n$$
we have
$$n \log \frac{n}{e} \lt \log n! \lt \log e+ n \log \frac{n}{2}$$
Proof:
For left side first step of induction is clear. Then
$$(n+1)!=n!(n+1) \gt \left( \frac{n}{e}\right)^n (n+1) =$$
$$=\left( \frac{n+1}{e}\right)^{n+1} \frac{(n+1)\left( \frac{n}{e}\right)^n}{\left( \frac{n+1}{e}\right)^{n+1}} \gt \left( \frac{n+1}{e}\right)^{n+1}$$
because $(n+1)\left( \frac{n}{e}\right)^n \left( \frac{n+1}{e}\right)^{-n-1}\gt 1$ is equivalent $\left(1+ \frac{1}{n}\right)^{n} \lt e$.
For right side
$$n! \lt \left(\frac{n+1}{2}\right)^{n} = e\left(\frac{n}{2}\right)^{n} \frac{\left(\frac{n+1}{2}\right)^{n}}{e\left(\frac{n}{2}\right)^{n}} = $$
$$=e\left(\frac{n}{2}\right)^{n} \frac{\left(1+ \frac{1}{n}\right)^{n}}{e} \lt e\left(\frac{n}{2}\right)^{n}$$
A: if we have $f(x) > 0$ and $f'(x) > 0,$  then
$$ \int_{a-1}^{b} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a}^{b+1} \;  f(x) \; dx   $$
$a=2; b= n ; f: log(n!)$
$$ \int_{1}^{n} \; \log x \;  dx \;  < \; \log n! \; = \; \sum_{j=2}^n \;  \log j \; < \; \int_{2}^{n+1} \;  \log x \; dx   $$
An antiderivative of $\log x$ is $x \log x - x$
$$ 1 - \frac{n-1}{n \log n} < \frac{\log n!}{n \log n} < \frac{(n+1) \log (n+1)}{n \log n} - \frac{n + 2 \log 2 -1}{n \log n}  $$
Some terms evidently go to zero with increasing $n.$
$$  \frac{(n+1) \log (n+1)}{n \log n} = \frac{(n+1) }{n } \frac{ \log (n+1)}{ \log n}   $$ goes to $1$
