Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove, for every $x>0$ $$\dfrac{1}{\ln^2\left(1+\dfrac{1}{x}\right)} \dfrac{1}{(x+1)x}-1>0.$$

share|improve this question
    
Does $ln^2()$ mean square of the logarithm or take the logarithm twice? –  Henry Feb 8 '12 at 8:45
    
Which operator is between first and second fraction ? –  pedja Feb 8 '12 at 8:45
    
@pedja multiple –  Joe Feb 8 '12 at 8:51
    
@Henry $(ln())^2$ –  Joe Feb 8 '12 at 8:52
    
Define $f(x)$ as : $$f(x)=\frac{1}{x(x+1)}-(\ln(x+1)-\ln x)^2$$ then calculate $f'(x)$ and try to show that $f'(x)>0$ for all $x>0$ . –  pedja Feb 8 '12 at 9:07

2 Answers 2

Here are some hints or ideas.

Exploring a few equivalent algebraic reformulations, for $x>0$, $$ \dfrac{1}{\ln^2\left(1+\dfrac1{x}\right)} \cdot \dfrac1{(x+1)x} - 1 > 0 $$ $$ \iff \ln^2\left(1+\dfrac1{x}\right) < \dfrac1{(x+1)x} = \dfrac1{x}-\dfrac1{x+1} $$ $$ \iff \ln\left(1+\dfrac1{x}\right) < \dfrac1{\sqrt{x(x+1)}} $$ $$ \iff \ln\left(1+u\right) < \dfrac{u}{\sqrt{u+1}} \quad \text{for} \quad u=\frac1x>0. $$ The last of these is fairly easy to work with. Setting $$ f(u)=\ln\left(1+u\right), \quad g(u)=\dfrac{u}{\sqrt{u+1}}=(1+u)^\frac12-(1+u)^{-\frac12} $$ we see that $f(u),g(u)\geq 0$ for $u>0$, that $f(0)=g(0)=0$, that $f(u),g(u)\approx u$ for $u\approx0$, and that $$ \lim_{u\rightarrow\infty}\frac{f(u)}{g(u)}=0 $$ (using L'Hôpital's rule) so graphing the two functions should get us almost there. For this, we probably only need the first one or two derivatives, $$ f'(u)=\left(1+u\right)^{-1}, \quad g(u) = \frac{ (1+u)^{-\frac12} - (1+u)^{-\frac32} }{2} = \frac12u(1+u)^{-\frac32} $$ which at zero are $$ f'(0)=g'(0)=1 \quad\text{and} $$ $$ f''(0)=-1,\quad g''(0)=-\frac12. $$ This should be enough to produce a convincing graph something like below,

enter image description here

made with sage (online).

var('u')
G = plot(log(1+u),      (u,0,10),color='red')
G+= text('f(u)',           (3,1),color='red')
G+= plot(u*(1+u)^(-1/2),(u,0,10),color='blue')
G+= text('g(u)',           (1,1),color='blue')
G.show()

For the full Taylor series (which we probably don't want), we would need the $n^\text{th}$ derivatives $$ f^{(n)}(u)=\frac{(-1)^{n-1}(n-1)!}{(1+u)^{n}} $$ and $$ g^{(n)}(u) =\frac12a_{n-1}(1+u)^{\frac12-n}-a_n(1+u)^{-\frac12-n} $$ for $n>0$, where $$ a_n =(-1)^n \left(\frac12\right) \left(\frac32\right) \cdots \left(\frac{2n-1}2\right) =(-1)^n \frac{(2n)!}{2^{2n}n!} =\left(n-\frac12\right)a_{n-1} $$ so that $$ g^{(n)}(u) =a_{n-1}\frac{\frac12(1+u)-\left(n-\frac12\right)}{(1+u)^{n+\frac12}} =a_{n-1}\frac{1+\frac{u}2-n}{(1+u)^{n+\frac12}}, $$ but there is probably an easier way than using the full Taylor series.

share|improve this answer

Define the function: $$f(x)=\frac{1}{ln^2(1+\frac{1}{x})}\frac{1}{x(1+x)}$$ The reciprocal of this function must be less than $1$. Note that $$\lim_{x=\infty}{f(x)^{-1}}=1$$ and that $$\lim_{x=0}{f(x)^{-1}}=0$$ Because the derivative of the $f(x)^{-1}$ is never zero, this means that $1$ is an asymptote because there isn't a maximum between zero and $+\infty$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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