# Radius & interval of convergence of power series

1. Oct 30, 2011

### kiwifruit

1. The problem statement, all variables and given/known data
i was doing this exercise and came across this example.

Ʃ (x^n)/ln(n+1)
n=1

3. The attempt at a solution
i know you have to do the ratio test which is
lim | a(n+1)/a(n)|
n>∞

i got to

lim | [x ln(n+1)] /ln(n+2) |
n>∞

and have no idea how to continue? is there a way to get rid of the logs? since in this case as n>∞ the top and bottom just goes to ∞?

Last edited: Oct 30, 2011
2. Oct 30, 2011

### susskind_leon

ln(n)/ln(n+1) goes to 1 for n -> infty
You can prove that easily by proving that ln(n)/ln(n+1)-1 approaches zero.

3. Oct 30, 2011

### HallsofIvy

Staff Emeritus
Or use L'Hopital's rule: the derivative of ln(x+1) is 1/(x+1) and the derivative of ln(x) is 1/x so this becomes (1/x)(x+1/1)= (x+1)/x goes to 1 as x goes to infinity.

That shows that ln(x)/ln(x+1) goes to 1 as x goes to infinity which includes n going to infinity.