Exercise 3 Let ax be a power series with radius of convergen

Exercise 3. Let ax be a power series with radius of convergence R > 0, prove that the radius of convergence of Sigma anx2 is 1.

Solution

Since {a_nx^n} converges, there is some positive real number B with the property that for all n one has

|a_nx^n| <= B.

It follows that for any positive integer n and any real x we have

|a_n x^n²| = |a_n| |x|^n² <= B |x|^n.

If x is a real number satisfying |x| < 1, the series sum B |x|^n² is a convergent geometric series; the above inequality and the comparison test then implies that sum |a_n x^n²| converges in that case, and the fact that absolute convergence implies convergence then tells us that sum a_n x^n² converges in that case.

So the given power series sum a_n x^n² converges for all x satisfying |x| < 1. This implies that the radius of convergence, R, of the series, satisfies R >= 1. But the fact that the series converges when x = 1 tells us that R <= 1. These two inequalities together imply that R = 1.