How to prove this inequality $(\frac{n+1}{e})^{n} < n! < e(\frac{n+1}{e})^{n+1}$?

Question

$\Bigl(\frac{n+1}{e}\Bigr)^{n} < n! < e\Bigl(\cfrac{n+1}{e}\Bigr)^{n+1}$

Answer 1

Both inequalities boil down to showing that for any $m \in \mathbb N$:

$$\int_{1}^{m} \ln(x)dx \ge \sum_{k=1}^{m-1} \ln(k)$$

And:

$$\int_{1}^{m} \ln(x) dx \le \sum_{k =1}^{m} \ln(k)$$

Both might be done by induction. The first one is easy to establish. For the second one, use:

• $$\int_{1}^{m} \ln(x) dx = \int_{1}^{m-1} \ln(x) dx + \int_{m-1}^{m} \ln(x)dx$$

• MVT for integrals.

• $\ln$ is increasing.

Answer 2

By Stirling formula, there exists a number $0<\theta<1$ such that $$n!=\sqrt{2\pi n}\left(\frac ne\right)^n e^{\frac \theta{12n}}.$$ So we have to prove that

$$\Bigl(\frac{n+1}{e}\Bigr)^{n} < \frac{1}{n+1}\sqrt{2\pi (n+1)}\left(\frac {n+1}e\right)^{n+1} e^{\frac \theta{12(n+1)}}< e\Bigl(\cfrac{n+1}{e}\Bigr)^{n+1},$$ that is

$$\frac{e}{n+1} <\frac{1}{n+1}\sqrt{2\pi (n+1)} e^{\frac \theta{12(n+1)}}< e,$$

which holds (for $n\ge 1$) because

$$e<3<\sqrt{4\pi}\le \sqrt{2\pi (n+1)}$$ and

$$\frac{1}{n+1}\sqrt{2\pi (n+1)} e^{\frac \theta{12(n+1)}}< \sqrt{\frac{2\pi}{n+1}} e^{\frac 1{24}}\le \sqrt{\pi} e^{\frac 1{24}}< e,$$ because of the following sequence

$$\pi^{\frac 12}<e^{\frac {23}{24}}$$

$$\pi^{12}<e^{20}$$

$$\pi^{3}<33<53<e^{4}<e^5.$$

Answer 3

We wish to show that $n!$ is bounded by the inequalities

$$\left(\frac{n+1}{e}\right)^{n}<n!<e\left(\frac{n+1}{e}\right)^{n+1}\tag1$$

We will prove these inequalities using mathematical induction.

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Let's start with the left-hand side inequality in $(1)$. First, we establish a base case. For $n=2$, we see that $2!>\left(\frac{3}{e}\right)^2$. Therefore, we assume that for some $n$, we have $\left(\frac{n+1}{e}\right)^{n}<n!$. Now, for $n+1$ we write

$$\begin{align} (n+1)!&=(n+1)n!\\\\ &>(n+1)\left(\frac{n+1}{e}\right)^{n}\\\\ &=\left(\frac{n+2}{e}\right)^{n+1}\,\left[e\left(\frac{n+1}{n+2}\right)^{n+1}\right]\\\\ &=\left(\frac{n+2}{e}\right)^{n+1}\,\frac{e}{\left(1+\frac{1}{n+1}\right)^{n+1}}\\\\ &>\left(\frac{n+2}{e}\right)^{n+1} \end{align}$$

where in the last inequality, we used $\left(1+\frac{1}{n+1}\right)^{n+1}<e$.

This concludes the proof by mathematical induction.

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Next, we analyze the right-hand side inequality in $(1)$. First, we establish a base case. For $n=2$, we see that $2!<e\left(\frac{3}{e}\right)^2$. Therefore, we assume that for some $n$, we have $e\left(\frac{n+1}{e}\right)^{n+1}>n!$. Now, for $n+1$ we write

$$\begin{align} (n+1)!&=(n+1)n!\\\\ &<(n+1)e\left(\frac{n+1}{e}\right)^{n+1}\\\\ &=e\left(\frac{n+2}{e}\right)^{n+2}\,\left[e\left(\frac{n+1}{n+2}\right)^{n+2}\right]\\\\ &=e\left(\frac{n+2}{e}\right)^{n+2}\,\left[e\left(1-\frac{1}{n+2}\right)^{n+2}\right]\\\\ &<e\left(\frac{n+2}{e}\right)^{n+2} \end{align}$$

where in the last inequality, we used $\left(1-\frac{1}{n+2}\right)^{n+2}<e^{-1}$.

This concludes the proof by mathematical induction.