Proof $(\frac{n+1}{n})^n>2$ for positive $n$ I would like to see some proofs that $(\frac{n+1}{n})^n>2$ for $n\in\mathbb R^+$ have some experience with inequalities, but I don't know too much theory.
Regards
 A: Your inequality fails to be true for real numbers $n$ in the interval $(0,1]$
However if you remove that condition and require $n>1\wedge n\in \mathbb{N}$
Then using the binomial theorem:
$$(1+\frac{1}{n})^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\binom{n}{0}\frac{1}{n^0}+\binom{n}{1}\frac{1}{n}+\sum_{k=2}^n\binom{n}{k}\frac{1}{n^k}=2+\sum_{k=2}^n\binom{n}{k}\frac{1}{n^k}>2$$
A: An alternate technique for arbitrary $n$:
$$[(1+1/n)^{n}]' = \log(1+1/n)\cdot(1+1/n)^{n} + n\cdot(1+1/n)^{n-1}\cdot (-1/n^2) \\
\geq \log(1+1/n)\cdot (1+1/n) - 1/n.$$
Now, consider $f(x) = \log(x+1)(x+1) - x$. Then, $f'(x) = \log(x+1) + 1 - 1 = \log(x+1) > 0$, $\forall x>0$.
Also, $f(0) = 0$. Hence, $f(x) > 0$, $\forall x>0$. So, in particular, our derivative is positive.
Now, evaluate $\lim_{n \rightarrow 0^{+}} (1+1/n)^{n}$, where $\frac{[\log(1+1/n)]'}{-1/n^2} = \frac{1}{1+1/n} \rightarrow 0$, whence $(1+1/n)^n > 1$.
But consider $n=1/2$. Then $(1+1/(1/2))^{1/2} = 3^{1/2} < 2$. If you assume $n\geq 1$, then, indeed, the value must be greater than or equal to $2$.
