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

1 Answer
dansmath
Sep 13, 2014

I used to live in Hicksville too, when I was a kid!

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 strikes again |

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