What are all values?

Question

What are all values of x for which the series ∑ (x-1)^n/n from n=1 to infinity converges?

 

Answer: 0≤x<2

 

Please show your work step by step. I'm new to this topic.

Answer 1

The ratio test only gives you the radius of convergence, from which you get an open interval, in this case (0,2). You will need to check the two interval endpoints for convergence separately. At x=2, the series is ∑(1/n), which is the harmonic series, which diverges, so x=2 is not in the interval of convergence. At x=0, the series is ∑(-1)ⁿ/n, which is the alternating harmonic series, which converges, so x=0 is in the interval of convergence. Therefore, the answer is 0≤x<2, or [0,2).

Answer 2

Use d"Alembert's ratio criteria according to which

 

    lim abs(a_n+1/a_n) < 1   where a_n = (x-1)ⁿ/n

    n→∞

 

By applying the criteria we will obtain

 

           lim  abs ((x-1) n/(n+1)) < 1

          n→∞

           

Because lim (n+1)/n = 1   this means     -1 < x-1 < 1

             n→∞            

 

 

or (by adding 1 to both sides)

          

 

                                     0<x<2