# Radius and Interval of convergence

1. May 10, 2013

### Painguy

1. The problem statement, all variables and given/known data
Ʃ(((-1)^n)(x^n))/(n+1) from 0 to ∞

2. Relevant equations

3. The attempt at a solution
I took applied the ratio test and got that lim|(x^(n+1))/(n+2) * (n+1)/(x^n)| =|x|
so that means for it to converge |x|<1 Radius of convergence is 1
My interval is (-1<x<1)
Now I check the endpoints

x=1
Ʃ(((-1)^n))/(n+1) from 0 to ∞
I want to use the alternating series test, but I have run into a little problem.
I know that the Lim 1/(n+1) =0 but how do I prove that 1/(n+2) < 1/(n+1) for all n.
I know its true, but my professor wants us to show that. I can show that -1<1 which is a true statement, but what does that say about the inequality in relation to n?

Is there a better test to do this? Ratio test just equals 1 so that tells me nothing. I can't really use the integral test for an alternating series since it would end up imaginary. If there is a better test could someone still show how to use the alternating series if possible?

2. May 11, 2013

### vela

Staff Emeritus
Usually you can assume certain properties of real numbers to be true. If your professor really expects you to prove such a trivial point, why don't you start with something "obviously" true, like n+2 > n+1 for all n?

3. May 11, 2013

### HallsofIvy

Staff Emeritus
Or, to start from the very obvious 2> 1. Add n to both sides, then divide both sides by ...

4. May 12, 2013

### Painguy

I set 1/(n+1) to a function f which I then siad was less than 0. when i took the derivative of f i obtained a negative function which is less than 0. I stated that since the derivative is less than zero then the function is decreasing and bn+1 < bn. Is that ok?