Would someone mind showing me how to do this proof?
Use Bernoulli's inequality to show that (1 + (1/n))^n < (1 + (1/(n+1)))^(n+1) for all n in N.
I'm given a hint of let x = -1/(n+1)^2.
Thank you in advance.
$\displaystyle \left( {1 - \frac{1}{{n + 1}}} \right)^{n + 1} \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} = \left( {1 - \frac{1}{{\left( {n + 1} \right)^2 }}} \right)^{n + 1} $
$\displaystyle \left( {1 - \frac{1}{{\left( {n + 1} \right)^2 }}} \right)^{n + 1} \ge 1 + \left( {n + 1} \right)\left( {\frac{{ - 1}}{{\left( {n + 1} \right)^2 }}} \right) = \left( {1 - \frac{1}{{n + 1}}} \right)$
$\displaystyle \left( {1 - \frac{1}{{n + 1}}} \right)^{n + 1} \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} \ge \left( {1 - \frac{1}{{n + 1}}} \right)$
$\displaystyle \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} \ge \left( {1 - \frac{1}{{n + 1}}} \right)^{ - n} = \left( {1 + \frac{1}{n}} \right)^n $
Yeah, sorry for the confusion. I meant the original inequality (the one proven using Bernoulli's Inequality). This one:
(1 + (1/n))^n < (1 + (1/(n+1)))^(n+1) for all n in N.
This is just a hint for how to attack the problem as given in the textbook.
Use x1 = 1, xk = (1 + (1/n)), and k = 2,3,...,n+1.
$\displaystyle \displaystyle\frac{1+\left(1+\frac{1}{n}\right)+\l eft(1+\frac{1}{n}\right)+\ldots+\left(1+\frac{1}{n }\right)}{n+1}>\sqrt[n+1]{\left(1+\frac{1}{n}\right)^n}$
$\displaystyle \displaystyle\frac{(n+1)+1}{n+1}>\left(1+\frac{1}{ n}\right)^{\frac{n}{n+1}}$
Rise both members to $\displaystyle n+1$.
$\displaystyle \left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1 }{n}\right)^n$