prove that for any $\displaystyle n$

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

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- Oct 22nd 2008, 07:46 AMgreat_mathInequality
prove that for any $\displaystyle n$

$\displaystyle 2\le\left(1+\frac{1}{n}\right)^n<3$ - Oct 22nd 2008, 02:50 PMPlato
Do you know BERNOULLI's INEQUALITY? $\displaystyle x > - 1 \Rightarrow \quad \left( {1 + x} \right)^n \geqslant 1 + nx$.

So $\displaystyle \left( {1 + \frac{1}{n}} \right)^n \geqslant 1 + n\left( {\frac{1}{n}} \right) = 2$.

That gives one ‘side’ of the inequality.

Using Napper inequality we can get $\displaystyle \ln \left( {1 + \frac{1}{n}} \right) < \frac{1}{n}$.

Using the exponential we get $\displaystyle \left( {1 + \frac{1}{n}} \right) < e^{\frac{1}

{n}} \Rightarrow \left( {1 + \frac{1}{n}} \right)^n < e < 3$ - Oct 31st 2008, 05:00 PMJes
What a slick way to get the lower bound. (Cool) I normally induct using the binomial expansion or the identity $\displaystyle a^n-b^n = (b^n + b^{n-2}a + \cdots + a^{n-1})$ to derive $\displaystyle b^n < a^n + (b-a)nb^{n-1}$ in order to show the sequence is monotonic increasing. Then it follows that $\displaystyle \inf \left (n + \frac{1}{n}\right )^n = 2$ since it is the first term. Let's see if I can use your method.

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

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