Find the radius of convergence?

Question

Find the radius of convergence of the power series ∑ x^n from n=0 to infinity. 

 

Answer: 1

 

I did the ratio test and got limit as n approaches to infinity of absolute value of x^(n+1) divided by x^n. But how is the answer 1? Please show all your work that leads to the answer. Thanks.

Answer 1

By ratio test,  L = lim(n -> inf) |x^(n+1) /x^n|  =  lim(n->inf)|x|  = |x|

         Therefore  L = |x|

The ratio test tells us that if L<1 the series will converge,  if L>1 the series will diverge, and if L=1 nobody knows what will happen.  

         Therefore  |x| < 1   series converge

                         |x| > 1   series diverge

       And the radius of convergence is 1

At x=1, 

       Sigma (n=0, infinity)1^n  diverge

At x =-1

        Sigma (n=0, infinity)(-1)^n diverge

Therefore,  the interval of convergence is then,     -1 < x < 1