<s>[INST] How to prove $(1+1/x)^x$ is increasing when $x>0$? Let $F(x)=(1+\frac{1}{x})^x$.
How do we prove $F(x)$ is increasing when $x>0$?
 [/INST] This proof is from theorem 140 from Hardy's Inequalities.
Let $f(x) = \ln\left[\left(1 + \frac{1}{x}\right)^x\right] = x(\ln(x+1) - \ln(x)).$ We refer to the mean value theorem: for each differentiable $g$,
$$
g(x + h) - g(x) = hg'(x + \theta h)
$$
for $\theta \in (0,1)$.
Applying the MVT to $g(x) = \ln(x)$, we get $\ln(x+1) - \ln(x) = \frac{1}{x + \theta}$. Since, $\frac{1}{x+\theta} > \frac{1}{x+1} $, we have that $\ln(x+1) - \ln(x) > \frac{1}{x+1}$, so
$$
f'(x) = \ln(x+1) - \ln(x) - \frac{1}{x+1} > 0.
$$
Hence, $f(x)$ and $e^{f(x)} = \left(1+\frac{1}{x}\right)^x$ are increasing with $x$.
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