I'm trying to use the mean value theorem to prove that
$$\frac{1}{n+1} < \ln(n+1)-\ln(n) < \frac{1}{n}$$
I've tried turning it into $\dfrac{1}{b} < \ln(b)-\ln(a) < \dfrac{1}{a}$ and then then dividing by $b-a$ but I'm not sure if that's even the right way to start.
Can someone please give me a hand?
Let $f(x)=\ln x$ then $f$ is continuous in $[n,n+1]$ and differentiable in $(n,n+1)$ for $n\in\mathbb{N}$. Mean value theorem says there exist a real number $c\in(n,n+1)$ such that $$f'(c)=\frac{f(n+1)-f(n)}{(n+1)-(n)}=\frac{\ln(n+1)-\ln(n)}{(n+1)-(n)}=\ln(n+1)-\ln(n)$$ but $f'(c)=\dfrac{1}{c}$ and from $n<c<n+1$ we have $$\frac{1}{n+1}<f'(c)<\frac{1}{n}$$ which gives $$\frac{1}{n+1}<\ln(n+1)-\ln(n)<\frac{1}{n}$$ Conclusion: $$\color{blue}{\left(1+\dfrac{1}{n}\right)^{n}<e<\left(1+\dfrac{1}{n}\right)^{n+1}}$$