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

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closed as off-topic by Did, LeGrandDODOM, Ahaan S. Rungta, user26857, graydad Feb 15 '15 at 20:00

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Induction.${}{}$ – P.. Feb 25 '13 at 11:16
Alternatively, I think there is a cheeky way to do this by considering the properties of the function $f(n)=\log_e(n)$. I shall type it up once I've got the details right, but you can look into that as well if you wish. – fixedp Feb 25 '13 at 11:35

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


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

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.

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How do you get to your first double inequality? I know you say it is by dividing by the lhs. – 1015 Feb 25 '13 at 11:50
There's a mistake in the left hand of my inequality. Thanks for the note, I'll try to fix it or else I'll erase my post. – DonAntonio Feb 25 '13 at 12:00
You could replace the first sentence by "Multiplying the whole thing by $e^n$". – Martin Feb 25 '13 at 12:34
Indeed, much simpler @Martin , thanks. – DonAntonio Feb 25 '13 at 12:37
+1 for thinking, arranging and long writing the answer. – Babak S. Feb 25 '13 at 14:24

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}

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Whoa we posted the same thing within minutes. – fixedp Feb 25 '13 at 12:53

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 $-(n+1)$)

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$\log x\,$ is increasing for all $\,x>0\,$ , its natural definition domain. – DonAntonio Feb 25 '13 at 13:00
Yeah, I suppose I just wanted to highlight that everything I did was for $x>1$. – fixedp Feb 25 '13 at 13:04
Thanks for checking. I was doing the integral from $x=1 $to $n$ in my working, but used the result for $n+1$ instead. I'll change it back to $n$ for convenience... – fixedp Feb 25 '13 at 13:15

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