Prove $\left(\frac{n+1}{\text{e}}\right)^n<n!<\text{e}\left(\frac{n+1}{\text{e}}\right)^{n+1}$ [closed]

Multiplying the whole thing through by $\,e^n\,$ and doing a little algebra we get an easier, imo, expression to prove:

$$(n+1)^n<e^nn!<(n+1)^{n+1}$$

$$n=1:\;\;\;\;\;\;2<e\cdot 1!<(2)^2\Longleftrightarrow 2<e<4\ldots\text{true.}$$

Now assume for $\,n\,$ and prove for $\,n+1\,$:

$$(n+1)^n<e^nn!<(n+1)^{n+1}\stackrel ?\Longrightarrow (n+2)^{n+1}<e^{n+1}(n+1)!<(n+2)^{n+2} $$

Let us focus on the right inequality:

$$e^{n+1}(n+1)!=e(n+1)e^nn!\stackrel{\text{ind. hypot.}}<e(n+1)(n+1)^{n+1}=e(n+1)^{n+2}$$

So it is enough now to prove that

$$e(n+1)^{n+2}<(n+2)^{n+2}\Longleftrightarrow e<\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{n+1}\right)\ldots$$

and I'll let the final touch to you (hint: remember the limit definition of the number $\,e\,$ ...)

For the left hand inequality: the same inductive assumption and

$$e^{n+1}(n+1)!=e(n+1)e^nn!\stackrel{\text{ind. hyp.}}>e(n+1)(n+1)^n=e(n+1)^{n+1}$$

so it's enough to show

$$e(n+1)^{n+1}>(n+2)^{n+1}\Longleftrightarrow e>\left(1+\frac{1}{n+1}\right)^{n+1}$$

Again, with the same hint as at the end of the first part, give the final touch here.

First denote, that $\log n! = \sum_{x=1}^n \log x$. Now we can simply bound the sum from above and below by: \begin{align} & \int_1^n \log (x) \, dx \leq \sum_{x=1}^n \log (x) \leq \int_0^n \log (x+1) \, dx \\ \Leftrightarrow & n \log(n )-n+1 \leq \log n! \leq (n+1)\log(n+1)-n \\ \Leftrightarrow & n \log(\frac ne )+1 \leq \log n! \leq (n+1)\log(\frac{n+1}{e})+1 \end{align} Now apply the exponential function to get \begin{align} & n \log(\frac ne )+1 \leq \log n! \leq (n+1)\log(\frac{n+1}{e})+1 \\ \Leftrightarrow &\Bigl(\frac ne\Bigr)^n e \leq n!\leq \Bigl(\frac{n+1}{e}\Bigr)^{n+1}e \end{align} This isn't exactly what you asked for, but consider the following: \begin{align} \Bigl(\frac{n+1}{e}\Bigr)^n =\Bigl(\frac{n}{e}\Bigr)^n\Bigl(\frac{n+1}{n}\Bigr)^n =\Bigl(\frac{n}{e}\Bigr)^n\Bigl(1+\frac{1}{n}\Bigr)^n\leq\Bigl(\frac{n}{e}\Bigr)^n e \end{align} Because $\Bigl(1+\frac{1}{n}\Bigr)^n$ is monotone increasing and converging towards $e$ (there are several proofs). Thus \begin{align} \Bigl(\frac{n+1}{e}\Bigr)^n\leq \Bigl(\frac{n}{e}\Bigr)^n e \leq n!\leq \Bigl(\frac{n+1}{e}\Bigr)^{n+1}e \end{align}

I was not able to complete the whole proof in detail, but here is a sketch anyway. There are some minor issues regarding what happens when $n =0,1$, but those two cases can be treated separately (and trivially). First a few observations

• $f(x)=\log_e(x)$ is an increasing function for $x>1$
• By considering rectangles of width 1 and height $f(x)$ from $x=1 $ to $n$, we see that $$f(1)+f(2)+...+f(n-1)<\int_1^{n} \log_e(x)dx<f(1)+f(2)+...+f(n) $$

Using basic log laws for the outer terms, and evaluating the integral in the middle, we obtain $$\log_e(n!)<\log_en^n-n+1<\log_e((n+1!))$$ Exponentiating and some rearrangements give $$(n-1)!<\frac{n^{n}}{e^{n-1}}<n!$$

This gives us the inequality on the right $n!<\frac{(n+1)^{n+1}}{e^n}$ when we replace $n$ with $n+1$. We also have $$\frac{n^{n}}{e^{n-1}}<n!$$ which is almost what we wanted, except that we need $\frac{(n+1)^n}{e^n}$ instead on the left hand side. It remains then to show that $$\frac{(n+1)^n}{e^n}<\frac{n^{n}}{e^{n-1}}\Leftrightarrow (n+1)<(e^{\frac{1}{n}}n)$$

This is easily seen to be true if we look at the series expansion of $(e^{\frac{1}{n}}n)-(n+1) $ (the first two terms of $(e^{\frac{1}{n}}n)$ cancels with $-(n+1)$)