calculus - Show that, for all $n > 1: frac{1}{n + 1} < log(1 + frac1n) < frac1n.$

I'm learning calculus, specifically derivatives and applications of MVT, and need help with the following exercice:

Show that, for all $n > 1$

$$\frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$$

I tried to follow the below steps in order to prove the RHS inequality:

Proving that $f < g$ on $I$ from $a$ to $b$:

Step $1$. Prove that $f' < g'$ on $\operatorname{Int}(I)$.

Step $2$. Show that $f(a) \leq g(a)$ or that $f(a^+) \leq g(a^+)$

Following the above steps, let $f(x) = \log(1 + \frac1x)$ and $g(x) = \frac1x$, for all $x > 1$. One has

$$f'(x) = -\frac{1}{x^2 + x} \quad \text{and} \quad g'(x) = -\frac{1}{x^2}.$$

We note that, for every $x > 1$, $f'(x) > g'(x)$. Moreover, $f(1^+) = \log(2) < 1 = g(1^+).$

My problem is that I got the wrong inequality sign in Step $1$.

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Looking at the solution in my textbook, the author suggests using the MVT but I don't know how to apply it in this case.

Answer

As suggested user84413 you can apply the MVT to $f(x)=\ln x$ on the interval $[n,n+1]$ for $n>0$: there is $t\in(n,n+1)$ such that

$$\log\left(1 + \frac1n\right)=\ln(n+1)-\ln(n)=f(n+1)-f(n)=f'(t)((n+1)-n)=\frac{1}{t}.$$

Now note that $0