Using the Mean Value Theorem, show that for all positive integers $n$, $n\ln(1+\frac{1}{n}) \leq 1$

Question

I am not too familiar with Rolle's Theorem and MVT so this question is a little bit tricky for me.

I tried it by letting some $$f(x)=x\ln(1+\frac{1}{x})-1.$$ Then by MVT, in the interval $x \in [0,n]$, there exists a point $c$, $c \in ]0,n[$, such that, $$\frac{f(n)-f(0)}{n-0} =f'(c)$$ Which simplifies to, $$\ln(1+\frac{1}{n})=f'(c).$$ By letting $c=0$ we get $$\ln(1+\frac{1}{n}) \leq f'(0)$$ Then $\ln(1+\frac{1}{n}) \leq \infty$ which is obviously nowhere near the desired inequality.

Answer 1

The inequality you really want is $$\ln(1+x)\le x\tag{1}$$ valid for $x>0$. Setting $x=1/n$ in ($1$) proves your given one.

To prove $(1)$, apply MVT to $f(t)=\ln(1+t)$.

Answer 2

Hint: $$\ln(1+\frac{1}{n})\leq \frac{1}{n} \Rightarrow 1+\frac{1}{n} \leq e^{\frac{1}{n}}$$

We know that $e = \lim_{n\to\infty}(1+\frac{1}{n})^n$ and we know $(1+\frac{1}{n})^n$ is increasing. Hence, for each $n$, $(1+\frac{1}{n})^n \leq e$ and $1+\frac{1}{n} \leq (1+\frac{1}{n})^n$.

Answer 3

$ \dfrac {\ln (1+1/n)-\ln 1}{1/n}= \ln' (t)= 1/t$,

where $1<t<1+1/n.$

Hence : $n \ln (1+1/n) =1/t \lt 1$.