How do you find the radius of convergence of a power series?

Question

William Abdalla · Accepted Answer

In my youth, I also resided in Hicksville!

To find the radius R of convergence of a power series #sum_(n=0)^(oo)c_n (x-a)^n,# centered at #x=a#, use the Ratio Test, and check that #lim_(n->oo) |(c_(n+1) (x-a)^(n+1))/(c_n (x-a)^n)|<1,# the same as #lim_(n->oo) |(c_(n+1))/(c_n)|*|x-a|<1,# or #|x-a|< lim_(n->oo) |(c_n)/(c_(n+1))|#

We wanted to find R such that our power series converged for #a-R < x < a+R#, which is #|x-a| < R#, so we use the value #R = lim_(n->oo) |(c_n)/(c_(n+1))|# for the radius of convergence.

Note: The word "radius" comes from the ability to use complex numbers for our variable x (and also the coefficients), and saying #|x-a| < R# describes the inside of a circle of real radius R in the complex plane, centered at the complex number a.

| Dansmath at it once more

Christopher Allers · Answer

undefined To find the radius of convergence ($R$) of a power series, you can use the Ratio Test. Here are the steps:

1. Consider a power series:
   $$ \sum_{n=0}^{\infty} a_n(x - c)^n $$

2. Apply the Ratio Test:
   - The Ratio Test states that if
     $$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$
     then the series converges absolutely if $ L < 1 $, diverges if $ L > 1 $, and the test is inconclusive if $ L = 1 $.

3. Calculate the limit:
   $$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

4. If the limit $ L $ exists, then the radius of convergence $ R $ is given by:
   $$ R = \frac{1}{L} $$
   
5. If the limit $ L $ does not exist or is infinite, then the radius of convergence $ R $ is $ R = \infty $.

So, to find the radius of convergence of a power series, calculate the limit $ L $ using the Ratio Test. If the limit exists and is not infinite, then $ R = \frac{1}{L} $. If the limit does not exist or is infinite, then $ R = \infty $.