Show $\left(1+\frac 1 x\right)^{x} < e < \left(1+\frac 1 {x}\right)^{x+1}$

Question

$\forall x > 0. \left(1+\frac 1 x\right)^{x} < e < \left(1+\frac 1 {x}\right)^{x+1}$

Unfortunately, I don't have this result yet in my theorem prover. What is an easy way to derive it?

Note that $e<\left(1+1/x\right)^{x+1}$ is equivalent to $\left(1-1/y\right)^y<e^{-1}$ where $y=x+1>1$. Not sure if that helps. — Thomas Andrews, May 29 '19 at 21:34
It all depends on what you have defined and proved for $e$ already. — Thomas Andrews, May 29 '19 at 21:35
Also note https://proofwiki.org/wiki/Lower_Bound_of_Natural_Logarithm — Calvin Khor, May 29 '19 at 22:06

AsdrubalBeltran · Accepted Answer · 2019-05-30T02:59:08.160

0

Note that $\left(1+\frac 1 x\right)^{x}=e^{x\log (1+1/x)}<e^{x\frac 1 x}=e$

$\left(1+\frac 1 x\right)^{x+1}=e^{(x+1)\log (1+1/x)}> e$

Edit

If $F(x)=(x+1)\log (1+1/x)$

$\lim_{x\to 0^+}F(x)=\infty$
$\lim_{x\to \infty}F(x)=\lim_{x\to \infty}\frac{\log (1+1/x)}{1/(x+1)}=1$, by L-hospital.
$F'(x)=\log (1+1/x)-1/x$, How $\log (1+1/x)<1/x\implies F'(x)<0$ then $F(x)$ is decreasing in $(0,\infty)$

Therefore $F(x)=(x+1)\log (1+1/x)>1$

edited May 30 '19 at 02:59

answered May 29 '19 at 21:43

AsdrubalBeltran

1

The first inequality is obvious, however, the last inequality needs to be proven. – dnqxt May 29 '19 at 21:53
@dnqxt ok now I proof this. – AsdrubalBeltran May 29 '19 at 22:33

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