# The asymptotic behavior of $n\ln n -n$ [closed]

How do I show that $$\displaystyle\lim_{n\to\infty}\dfrac{n\ln n - n}{\ln n!}=1?$$

• $\ln n! \le \ln n^n = n\ln n$, and for any $\epsilon > 0$, for $n$ large, $\ln n! \ge \ln [(n(1-\epsilon))^{n(1-\epsilon)}] \ge n(1-\epsilon)^2 \ln(n)$. – mathworker21 May 25 '19 at 23:59
• Have you learned about integrals yet? – saulspatz May 26 '19 at 1:15
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Using Stolz–Cesàro theorem you get \begin{align} \lim_{n\to\infty} \frac{n\ln n - n}{\ln n!} &= \lim_{n\to\infty} \frac{1 + (n+1)\ln(n+1) - n\ln n}{\ln (n+1)}\\ & = \lim_{n\to\infty} \frac{\ln(n+1) + n(\ln(n+1)-\ln n)}{\ln (n+1)} \\ &= \lim_{n\to\infty} \frac{\ln(n+1) + \ln\left(1+\frac1n\right)^{n}}{\ln (n+1)} \\ &= 1 \end{align}

You can also check some similar older questions (it seems natural that subtracting $$n$$ in the numerator does not change the limit):

don't need Stirling. For a function such as logarithm with $$f(x) > 0$$ and $$f'(x) > 0,$$ we get $$\int_{a-1}^b \; f(x) dx < \sum_{k=a}^b \; f(k) < \int_{a}^{b+1} \; f(x) dx$$

Here $$f$$ is log base e, take $$a=2$$ and $$b=n$$ $$\int_{1}^n \; \log x \; dx < \sum_{k=2}^n \; \log k < \int_{2}^{n+1} \; \log x \; dx$$

An antiderivative of $$\log x$$ is $$x \log x - x.$$

$$n \log n - n + 1 < \log n! < (n+1) \log (n+1) - n - 1 - 2 \log 2 + 2$$ Hint:

By Stirling' formula, $$\log n!\sim_\infty \log(2\pi n)+n\log n-n$$ so all you have to prove is that $$\frac{\log(2\pi n)}{n\log n-n}\to 0\quad\text{ as }\enspace n\to\infty.$$

• It seems to me that this problem is equally difficult. – Abo Rakan May 26 '19 at 1:08
• Just expand the numerator to log(2 Pi)+log(n) and then it all works out. – plus1 May 26 '19 at 2:22

As Bernard answered, the problem of the limit itself is quite simple using Stirling approximation of $$\log(n!)$$.

If you take one more term of this approximation, you could even get a quite good approximation since it will give $$\dfrac{n\ln n - n}{\ln n!}\sim 1-\frac{\ln (2 \pi n)}{2 n(\ln (n)-1)}$$ For example, using $$n=100$$, the exact value would be $$0.991141$$ while the approximation would give $$0.991064$$.