1
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ thanks so much! My mistake was that I was taking 1 as my lower bound instead of 0... $\endgroup$ – Cyril Gliner Feb 15 '17 at 1:38
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.