Proof

**Snooks02** · November 30th 2008, 05:40 PM

if f(1) + f(2) +... + f(n-1)< the integral from 1 to n of f(x)dx < f(2) + f(3) +...+ f(n).
Chose f(x) = ln x. Show that n^n/e^(n-1)< n! < n^(n+1)/e^n

**Mathstud28** · November 30th 2008, 06:39 PM

Originally Posted by Snooks02

if f(1) + f(2) +... + f(n-1)< the integral from 1 to n of f(x)dx < f(2) + f(3) +...+ f(n).
Chose f(x) = ln x. Show that n^n/e^(n-1)< n! < n^(n+1)/e^n

Assume that $(n!)'>0$ and $n>0$

$\begin{aligned}\sum_{k=1}^{n-1}\ln(k)&=\ln(1)+\ln(2)+\cdots+\ln(n-1)\\<br /> &=\ln\left(1\cdot{2}\cdots{n-1}\right)\\<br /> &=\ln((n-1)!)\end{aligned}$

And

$\int_1^n\ln(x)dx=n\ln(n)-n+1=\ln\left(\frac{n^n}{e^{n-1}}\right)$

As well as

$\begin{aligned}\sum_{k=2}^{n}\ln(k)&=\ln(2)+\ln(3) +\cdots+\ln(n)\\<br /> &=\ln\left(1\cdot{2}\cdots{n}\right)\\<br /> &=\ln(n!)\end{aligned}$

So we would have that

$\ln((n-1)!)<\ln\left(\frac{n^n}{e^{n-1}}\right)<\ln(n!)$

And since the exponential function is stricly increasing across its domain the above inequality also gives

$(n-1)!<\frac{n^n}{e^{n-1}}<n!$

So then because $n!$ is strictly increasing we may rewrite our inequality as

$1<\frac{n^n}{e^{n-1}}\cdot\frac{1}{(n-1)!}<n$

Reciporcating and not forgetting to reverse the direction of the inequalities gives

$\frac{1}{n}<(n-1)!\cdot\frac{e^{n-1}}{n^n}<1$

Multiplying through by $\frac{n^n}{e^{n-1}}$ gives

$\frac{n^{n-1}}{e^{n-1}}<(n-1)!<\frac{n^n}{e^{n-1}}$

Finally multiplying through by $n$ gives

$\frac{n^n}{e^{n-1}}<n!<\frac{n^{n+1}}{e^{n-1}}\quad\blacksquare$

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