# Using the Mean Value Theorem to prove an inequality

Can someone help me to solve this question:

Using the Mean Value Theorem, show that for all positive integers n:

$$n\ln{\big(1+\frac{1}{n}}\big)\le 1.$$

I've tried basically every function out there, and I can't get it. I know how to prove it using another technique, but how do you do it using MVT?

Thank you very much in advance,

C.G

## 3 Answers

Let $f(x)=\ln(1+x)$, then $f^{\prime}(x)=\frac{1}{1+x}$, hence by the mean value theorem for any $x>0$ there is some $0<t<x$ such that $$\frac{f(x)-f(0)}{x-0}=f^{\prime}(t)=\frac{1}{1+t}$$ Since $f(0)=0$ and $\frac{1}{1+t}<1$, this implies that $$\frac{f(x)}{x}<1$$ for all $x>0$, hence $$\ln(1+x)=f(x)<x$$ for all $x>0$. Now taking $x=\frac{1}{n}$ we get $$\ln\Big(1+\frac{1}{n}\Big)<\frac{1}{n}$$ for all $n\geq 1$, which is the desired result.

• thanks so much! My mistake was that I was taking 1 as my lower bound instead of 0... – Cyril Gliner Feb 15 '17 at 1:38

HINT: Write this as $$\frac{\ln(1+\frac1n)-\ln(1)}{\frac 1n}.$$

Let $f(x)=\ln x$ on interval $[n,n+1]$ for all $n\in\mathbb{N}$. $f$ is an increasing function with $f'(x)=\dfrac{1}{n}$, using the Mean Value Theorem, $$f(n+1)-f(n)=f'(c)(n+1-n)=\frac1c$$ for a $c\in[n,n+1]$, hence $$n<c<n+1$$ $$\frac{1}{n+1}<\frac1c<\frac1n$$ $$\frac{1}{n+1}<f(n+1)-f(n)<\frac1n$$ $$\frac{1}{n+1}<\ln\frac{n+1}{n}<\frac1n$$ with right inequality $$n\ln{\big(1+\frac{1}{n}}\big)< 1$$ for all positive integers $n$.