Prove $ 2<(1+\frac{1}{n})^{n}$ How to prove that $ 2<(1+\frac{1}{n})^{n}$ for every integer $n>1$. I was thinking by induction, it works for $n=2$ but then I couldn't move forward.
 A: Hint: Use the binomial theorem, look at the first two to three terms.
A: you can use  Bernoulli's inequality

for every $ x>-1$ $$\\ { \left( 1+x \right)  }^{ n }\ge 1+nx$$

A: To show $\Big( 1+\frac{1}{n}\Big)^n>2\Longleftrightarrow$ To show $(n+1)^n>2n^n$
Note that, for $n>1$, $$(n+1)^n=\sum_{k=0}^{n}\binom{n}{k}n^k>n^n+\binom{n}{1}n^{n-1}=n^n+n^n=2n^n\space\space\space\blacksquare$$
A: A Calculus approach: Show that the expression is greater than $2$ for $n=2$ $$\left.\left(1+\frac 1 n\right)^n\right|_{n=2}=\frac 9 4 > 2$$ then show that the expression is only increasing past $n=2$
$$ \frac{d}{dn}\left[ \left(1+ \frac 1 n \right)^n \right] = \frac{1}{n+1} \left( 1 + \frac 1 n \right)^n \left((n+1)\log\left(1 + \frac 1 n \right) - 1 \right) > 0 \quad \forall n $$
A: Bernoulli's Inequality says
$$
\begin{align}
\frac{\left(1+\frac1n\right)^n}{\left(1+\frac1{n-1}\right)^{n-1}}
&=\left(1-\frac1{n^2}\right)^n\frac{n}{n-1}\\
&\ge\left(1-\frac1n\right)\frac{n}{n-1}\\[9pt]
&=1
\end{align}
$$
Thus, $\left(1+\frac1n\right)^n$ is increasing and since $\left(1+\frac12\right)^2=\frac94\gt2$, we get the desired result.
A: Approach using AM-GM Inequality
First we show that 
$$f(n)=\left(1+\frac{1}{n}\right)^n$$
is monotone increasing. To see this, apply the AM-GM inequality to the following $n+1$ terms
$$\left\{1,1+\frac{1}{n},1+\frac{1}{n},\cdots n\mbox{ times}\right\}$$
You get
$$\frac{1+n+1}{n+1} \geq \left(1+\frac{1}{n}\right)^{\frac{n}{n+1}}$$
Rearranging gives you $f(n) < f(n+1)$ since the terms in AM-GM are not equal.
Since $f(1)=2$, the result follows.
A: By the Binomial theorem,
$$\left(1+\frac1n\right)^n=1+\binom n1\frac1n+\cdots=1+\frac nn+\cdots$$ where the extra terms are positive.
