Convergence of a sum for which x?

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Homework Statement


Consider the infinite series (1/n) * (xn) where x is a real noumber. Find all numbers x such that
i) the series converges,
ii) series converges absolutely
iii) diverges to + infinity
iiii) does not converge.




2. The attempt at a solution
For this, i know it does not converge for x = 1 and for x<-1 , and it does converge absolutely for x = -1, i also know it will diverge to \infty for x>1 but i am very unsure as to what happens for x between 1 and - 1 ?

Thanks a lot
 
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What does the ratio test tell you?
 
Ok, I see, so now i have that it converges for |x|<1 and the same for absolute convergence.
It does not converge at x=1 , x=-1 pr x<-1
and it diverges to infinity for x>1.
It this closer?
 
stukbv said:
Ok, I see, so now i have that it converges for |x|<1 and the same for absolute convergence.
It does not converge at x=1 , x=-1 pr x<-1
and it diverges to infinity for x>1.
It this closer?

Closer. It converges for |x|<1 and diverges for |x|>1. The only points you have to worry about are x=1 and x=(-1). I'll agree it diverges at x=1. Can you say why? I don't agree that it diverges at x=(-1).
 
For x= -1, this is an alternating series and 1/n goes to 0.
 
Oh yes, of course!
Thanks!
 
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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