Limit of ln n/ln (n+1) as n approaches infinity | Calculus Help and Solutions

[juliusoh]

Homework Statement

Find the limit of: lim as n--> ∞ of [ln n / ln (n+1)]^n

Homework Equations

The Attempt at a Solution

I used the Lopitals rule to find the limit of ln/ln(n+1) and it equals to 1.
However what do i do after there?
AND I know its bounded by 0< <1
squeeze theorem..
BUT IM not sure what to do next

 

[SammyS]

Homework Statement

Find the limit of: lim as n--> ∞ of [ln n / ln (n+1)]^n

Homework Equations

The Attempt at a Solution

I used the Lopitals rule to find the limit of ln/ln(n+1) and it equals to 1.
However what do i do after there?
AND I know its bounded by 0< <1
squeeze theorem..
BUT I'm not sure what to do next.

Hello juliusoh. Welcome to PF !

Try to find the limit of the log of your expression.

\displaystyle \lim_{n\to\infty} \ln\left(\left(\frac{\ln(n)}{\ln(n+1)}\right)^n\ \right)

 

[juliusoh]

Wow its too complicated.. i keep getting stuck...

 

[Bohrok]

This is a good problem to get you acquainted with l'Hopital's rule

Is this how you started it? I haven't worked it out, but this is how I would rewrite it before l'Hopital's rule:

\ln \left(\left(\frac{\ln x}{\ln(x+1)}\right)^x\right) = x\ln\left(\frac{\ln x}{\ln(x+1)}\right) = \frac{\ln\left(\frac{\ln x}{\ln(x+1)}\right)}{\frac{1}{x}}

 

[Dick]

If you are just interesting in finding the limit and less interested in a rigorous proof, you could do it the physicists way. Use approximations like log(1+x)~x and 1/(1+x)~1-x where x<<1 and the '~' means I've left out higher order terms in the taylor expansion. Start by writing log(n+1)=log(n*(1+1/n))=log(n)+log(1+1/n)~log(n)+1/n.

 

[juliusoh]

I see!
Thanks BohRok..
I don't know how to do the physics way, i haven't learned that yet. I am just a freshman in college.
So derivative of ln(ln x/ln(x+1)... how do you do that lol.

 

[Bohrok]

d/dx ln(f(x)) = f'(x)/f(x), then let f(x) = ln x/ln(x+1). Be careful when using the quotient rule to find f'(x)!