Discover Radius of Convergent for the Series: (1-2^n)(ln(n))x^n - Solved!

[danni7070]

[Solved] Radius of Convergent

Homework Statement

Find the radius of convergent for \sum_{n=1}^\infty (1-2^n)(ln(n))x^n

Homework Equations

\frac {1}{R} = L = \lim \frac{a_{n+1}}{a_n}

The Attempt at a Solution

lim \frac {(1-2^{n+1})(ln(n+1)}{(1-2^n)(ln(n))} = L

lim \frac {(1-2^n)(ln(n))}{(1-2^{n+1})(ln(n+1))} = R

I'm dizzy looking at this but how can I find:

\lim_{n\rightarrow\infty} R

 

[Mr.Brown]

my idea would be to take the quotient of the logs as 1 because the derivatives both go as 1/x and so both functions should behave the same in infinity.
The ones up front can be thrown away don´t make any contribution at infinity so you´re left with 2^n/2^n+1 = 1/2 = R :)

No make it a bit more rigorous if you like :)

 

[danni7070]

lim\frac{1-2^n}{1-2^{n+1}} * \frac{ln(n)}{ln(n+1)} = \frac{2^n}{2^{n+1}} = 2^{n-(n+1)} = 2^{-1} = \frac{1}{2}

Indeed this is right, I've checked the results and it is a hit!

But I don't underssand why you can just skip the log parts?

Ahh, now when I think about I see that lim \frac{ln(n)}{ln(n+1)} = 1 so that cancels out!

I usually get stuck with the obvious.

Thanks mr. Brown!

 

[Mr.Brown]

you could use l´hospital on the logs to make it rigorous be the derivative quotient would be
n^-1/(n+1)^(-1)=1+1/n goes to n as n goes to infinity :)

 

[danni7070]

Yeah I know but this lim is just so much for the eye to solve :)