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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.

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up vote 7 down vote accepted

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 $1.$

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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 $$0\geq \frac 1{n\log n}\sum_{j=1}^n\log\frac jn\geq \frac 1{n\log n}\sum_{j=1}^n\left(\frac jn-1\right)-\frac 1{2n\log n}\sum_{j=1}^n\frac{j^2}{n^2}.$$ We have in the RHS $\frac 1{\log n}$ times a Riemann sum (associated with $t\mapsto t-1-t^2/2$), hence $\frac 1{\log n}$ times a bounded term, which gives the result.

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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: $\log(n!)=n\log(n)-n + o(n)$.

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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.

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