<s>[INST] How to prove $\ln{\frac{n+1}{n}}\le\frac{2}{n+1},\forall n\in\mathbb{N}^+$? I found an inequality: $$\ln\left(\frac{n+1}{n}\right)\le\frac{2}{n+1},\forall n\in\mathbb{N}^+.$$
I tried induction. It is obvious if $k=1$, when $n=k$, $\ln\sqrt{\left(\frac{k+1}{k}\right)^{k+1}}\le 1$, but bogged down for $n=k+1$:
$$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$
 [/INST] Integrate $\frac{1}{x}$ between $n$ and $n+1$: $\ln \frac{n+1}{n} = \ln (n+1) - \ln (n) = \displaystyle\int_n^{n+1} \frac{1}{x} \operatorname{d}x$. As $x\to \frac{1}{x}$ is decreasing, you get $\ln \frac{n+1}{n}\leq \displaystyle\int_n^{n+1} \frac{1}{n} \operatorname{d}x = \frac{1}{n}$.
It is moreover clear that $\frac{1}{n} \leq \frac{2}{n+1}$ for $n\geq 1$, so you get $\ln \frac{n+1}{n} \leq \frac{2}{n+1}$
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