1. Dec 8, 2007

### danni7070

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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:

$$\lim_{n\rightarrow\infty} R$$

Last edited: Dec 8, 2007
2. Dec 8, 2007

### 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 :)

3. Dec 8, 2007

### 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!

4. Dec 8, 2007

### 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 :)

5. Dec 8, 2007

### danni7070

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