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.

I need to find

$\lim_{x\rightarrow0}f(x)$ for the following function:

$f:(0,+\infty)$

$f(x)=[1+\ln(1+x)+\ln(1+2x)+\dots+\ln(1+nx)]^\frac{1}{x}$

I tried writing the logarithms as products:

$\lim_{x\rightarrow0}[1+\ln(1+x)(1+2x)\dots(1+nx)]^\frac{1}{x}$

and as a sum and nothing is getting me anywhere.

Also I know I have to use the formula: $\lim_{x\rightarrow0}(1+x)^\frac{1}{x}=e$

Can someone please help me?

Thank you very much!

share|improve this question
    
You want $\lim_{x\to0}$ or $\lim_{n\to\infty}$? –  Julián Aguirre Jul 20 '12 at 9:04
    
I need $lim_{x\rightarrow0}$. Could you point me in the right direction? –  Grozav Alex Ioan Jul 20 '12 at 9:06
1  
@GrozavAlexIoan $\log(1+x)=x+o(x)$, $x\to0$. –  Andrew Jul 20 '12 at 9:08
    
I tried that. Unfortunately $e$ is not one of the possible answers. –  Grozav Alex Ioan Jul 20 '12 at 9:16
    
@Andrew's hint works all right (and does not yield the limit e). –  Did Jul 20 '12 at 9:21
show 2 more comments

2 Answers

up vote 1 down vote accepted

Here is the solution, take $\ln$ to both sides, gives,

$$\ln \left( f \left( x \right) \right) ={\frac {\ln \left( 1+\sum _{ k=1}^{n}\ln \left( 1+kx \right) \right) }{x}} $$

Using Taylor expansion of $\ln(1+t)= t+ O(t) $ at the point $t=0$ with $t = {\sum _{k=1}^{n}\ln \left( 1+kx \right) } $ yields

$$ \ln \left( f \left( x \right) \right) ={\frac {\sum _{k=1}^{n}\ln \left( 1+kx \right) }{x}} + \frac{O\left(\left( {\sum _{k=1}^{n}\ln \left( 1+kx \right) } \right)^2\right)}{x} $$

Taking the limit as x goes to $0$ to both sides of the above equation gives

$$ \lim_{x->0}\ln(f(x))=\ln(\lim_{x->0} f(x) ) =\sum _{k=1}^{n}{k}^{} =\frac{n(n+1)}{2}\,.$$

Exponentiating the last result, we get the answer

$$ {\rm e}^{\frac{n(n+1)}{2}} $$

share|improve this answer
    
There is an error. Check the link between the second equation and the third: in the third, no inverses should appear, so the limit is much simpler. –  D. Thomine Jul 20 '12 at 11:18
    
If $x$ yields $0$ wouldn't the result be $ln(f(x))=\sum^n_{k=1}0+O(\sum^n_{k=1}0)$ meaning $f(x)=e^0$ => $\lim f(x)=1$ ? –  Grozav Alex Ioan Jul 20 '12 at 11:35
    
Thanks for the correction. –  Mhenni Benghorbal Jul 20 '12 at 12:54
add comment

(Edited in response to comment)

$$\log f={\log\bigl(1+\log(1+x)+\log(1+2x)+\cdots+\log(1+nx)\bigr)\over x}$$ By l'Hopital, the limit, if it exists, is the same as the limit of $${\left({1\over1+x}+{2\over1+2x}+\dots+{n\over1+nx}\right)\over\left(1+\log(1+x)+\log(1+2x)+\cdots+\log(1+nx)\right)}$$ But now you can just set $x$ equal to zero.

share|improve this answer
    
Can you please explain what you did at the first part (how the $\frac{log}{x}$ appeared) and how you got the $(\frac{1}{1+x}+\frac{2}{1+2x}+\dots+\frac{n}{1+nx})$? Thank you very much! –  Grozav Alex Ioan Jul 20 '12 at 9:34
    
I took the log on both sides - are you familiar with $\log a^b=b\log a$? Are you familiar with l'Hopital? Do you know that the derivative of $\log(1+kx)$ is $k/(1+kx)? –  Gerry Myerson Jul 20 '12 at 9:40
    
I'm familiar with the logarithm properties and l'Hopital. I didn't know the derivative of $log(1+kx)$ though. This means the answer is $e^{\frac{n(n+1)}{2}}$m right? –  Grozav Alex Ioan Jul 20 '12 at 9:47
    
Note that the term in my answer that comes out to $n(n+1)/2$ is in the denominator. –  Gerry Myerson Jul 20 '12 at 10:48
    
@GerryMyerson Why is that term in the denominator? Isn't that term essentially your du, where u is equal to that sum with all the logs? –  Mike Jul 20 '12 at 11:11
show 1 more comment

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.