TITLE: Show that, for all $n > 1: \frac{1}{n + 1} < \log(1 + \frac1n) < \frac1n.$
QUESTION [9 upvotes]: 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 $1$.

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.

REPLY [2 votes]: For any $x>0$ We have 
$$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1}  \le \int_{0}^x\frac{dt}{1}  = x $$
that is 
$$\frac{x}{x+1} \le \ln(x+1 ) \le   x $$
taking $x=\frac1n$ you get 
$$\frac{1}{n+1} \le \ln(\frac1n+1 ) \le   \frac1n $$