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 $n=k+1$: $$\ln\sqrt{\left(\frac{k+2}{k+1}\right)^{k+2}}\le 1$$

Take $x=\frac1n$ so you will have $$\ln(1+x)\leq \frac{2x}{x+1}$$ now suppose $f(x)=\ln(1+x)- \frac{2x}{x+1}\\\x \in [0,1]$ $$f'(x)=\frac{1}{x+1}-\frac{2}{(x+1)^2}=\frac{1+x-2}{(x+1)^2} \to x=1$$ so $f'(x) \leq 0 \text { when x is } \in [0,1]$ so $f(x)$ is decreasing in this interval so $$f(x)\leq f(0) \text{ when } x\in [0,1]\\\ \to f(x)\leq f(0)\\\f(x) \leq \ln(1+0)-0\\\f(x)\leq 0\\\\\ln(1+x)- \frac{2x}{x+1}\leq 0\\\ \ln(1+x)\leq \frac{2x}{x+1} , x \in [0,1] \\\and \\\\\ln(1+x)\leq \frac{2x}{x+1} , x \in (0,1]$$ now put $x=\frac1n$