analysis inequality

**ochoCinco** · September 16th 2007, 12:23 PM

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.

**shilz222** · September 16th 2007, 12:31 PM

$(1+x)^r \geq 1+rx$

$\left(1 + \frac{1}{n}\right )^n < \left (1+ \frac{1}{n+1}\right )^{n+1}$

$x = -\frac{1}{(n+1)^2}$

$\left(1 - \frac{1}{(n+1)^2} \right)^n \geq 1-\frac{n}{(n+1)^2}$

**ochoCinco** · September 16th 2007, 12:38 PM

Well what does this mean?

How do you manipulate the inequality at the end?

**Plato** · September 16th 2007, 12:56 PM

$\left( {1 - \frac{1}{{n + 1}}} \right)^{n + 1} \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} = \left( {1 - \frac{1}{{\left( {n + 1} \right)^2 }}} \right)^{n + 1}$

$\left( {1 - \frac{1}{{\left( {n + 1} \right)^2 }}} \right)^{n + 1} \ge 1 + \left( {n + 1} \right)\left( {\frac{{ - 1}}{{\left( {n + 1} \right)^2 }}} \right) = \left( {1 - \frac{1}{{n + 1}}} \right)$

$\left( {1 - \frac{1}{{n + 1}}} \right)^{n + 1} \left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} \ge \left( {1 - \frac{1}{{n + 1}}} \right)$

$\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 1} \ge \left( {1 - \frac{1}{{n + 1}}} \right)^{ - n} = \left( {1 + \frac{1}{n}} \right)^n$

**ochoCinco** · September 16th 2007, 01:55 PM

Is there a similar way to prove this inequality using the Arithmetic-Geometric Mean Inequality?

Use x1 = 1, xk = (1 + (1/n)), and k = 2,3,...,n+1.

**Plato** · September 16th 2007, 02:02 PM

Please review what you posted!
It is not an inequality.
In fact, it is hard to know what you may mean.

**ochoCinco** · September 16th 2007, 02:06 PM

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.

**Plato** · September 16th 2007, 02:11 PM

The problem with hints is that often the only one who knows how it should be used is the person giving the hint. In this case, I have no idea.

**red_dog** · September 17th 2007, 09:16 AM

$\displaystyle\frac{1+\left(1+\frac{1}{n}\right)+\l eft(1+\frac{1}{n}\right)+\ldots+\left(1+\frac{1}{n }\right)}{n+1}>\sqrt[n+1]{\left(1+\frac{1}{n}\right)^n}$
$\displaystyle\frac{(n+1)+1}{n+1}>\left(1+\frac{1}{ n}\right)^{\frac{n}{n+1}}$
Rise both members to $n+1$ .
$\left(1+\frac{1}{n+1}\right)^{n+1}>\left(1+\frac{1 }{n}\right)^n$

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