Radius and Interval of convergence

Painguy
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Homework Statement


Ʃ(((-1)^n)(x^n))/(n+1) from 0 to ∞

Homework Equations


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?
 
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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?
 
Or, to start from the very obvious 2> 1. Add n to both sides, then divide both sides by ...
 
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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