Proof that $(1+1/x)^x$ is monotonic increasing How does one prove that $(1+\frac{1}{x})^x$ is monotonic increasing for any $x \in [1,\infty)$?
Thanks a million!
 A: $$f(x) = \left ( 1+ \frac{1}{x} \right )^x$$
$$\log{f(x)} = x \log{\left( 1+ \frac{1}{x} \right )}$$
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{x}{1+\frac{1}{x}}\left (\frac{-1}{x^2} \right ) = \log{\left( 1+ \frac{1}{x} \right )} - \frac{1/x}{1+ \frac{1}{x}}$$
or
$$\frac{d}{dx}\log{f(x)} = \log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1 $$
Because $y \log{y} \ge y-1 \; \forall y \ge 1$, we may say that, $\forall x \ge 1$:
$$\frac{d}{dx}\log{f(x)} =\log{\left( 1+ \frac{1}{x} \right )} + \frac{1}{1+ \frac{1}{x}} - 1  \ge 0$$
and $f(x)$ is monotone increasing for all $x \ge 1$, as $\log$ is monotone increasing.
A: Hint: Consider the sign of the derivative!
A: For $x\in \mathbb{N}$ this result can be derived using the AM-GM inequality:
$$(\frac{n.\frac{n+1}{n}+1}{n+1})\geq((\frac{n+1}{n})^n.(1)^{1})^{(\frac{1}{n+1})}$$
Using A.M. G.M. inequality(i.e A.M$\geq$G.M.)
$$\Rightarrow\frac{(n+1)+1}{n+1}\geq(\frac{n+1}{n})^{\frac{n}{n+1}}$$
$$\Rightarrow(1+\frac{1}{n+1})^{n+1}\geq(1+\frac{1}{n})^{n}$$
