Inequality

great_math · #1 $Old$ 10-22-2008, 09:46 AM

prove that for any $n$

$2\le\left(1+\frac{1}{n}\right)^n<3$

Plato · #2 $Old$ 10-22-2008, 04:50 PM

Do you know BERNOULLI's INEQUALITY? $x > - 1 \Rightarrow \quad \left( {1 + x} \right)^n \geqslant 1 + nx$ .
So $\left( {1 + \frac{1}{n}} \right)^n \geqslant 1 + n\left( {\frac{1}{n}} \right) = 2$ .
That gives one ‘side’ of the inequality.

Using Napper inequality we can get $\ln \left( {1 + \frac{1}{n}} \right) < \frac{1}{n}$ .
Using the exponential we get $\left( {1 + \frac{1}{n}} \right) < e^{\frac{1} {n}} \Rightarrow \left( {1 + \frac{1}{n}} \right)^n < e < 3$

Jes · #3 $Old$ 10-31-2008, 07:00 PM

What a slick way to get the lower bound.

I normally induct using the binomial expansion or the identity $a^n-b^n = (b^n + b^{n-2}a + \cdots + a^{n-1})$ to derive $b^n < a^n + (b-a)nb^{n-1}$ in order to show the sequence is monotonic increasing. Then it follows that $\inf \left (n + \frac{1}{n}\right )^n = 2$ since it is the first term. Let's see if I can use your method.

Let $a_n$ denote the sequence and let $M = \inf (a_n)$ . Assume $M$ is such that $2 < M \leq \left (n + \frac{1}{n}\right )^n$ . Then $a_1 = 2 < M$ contradicts that $M$ is a lower bound.

Next assume $M < 2.$ There is some $a_i$ in the sequence such that $M \leq a_i < 2$ otherwise $2$ is the greatest lower bound. But no such $a_i$ exists since it was proved that $2 \leq a_n < 3.$ By Trichotomy, $\inf (a_n) = 2.$