Thread: Multiplicative comutativity of power series

A(x) = sum(n=0 -> infinity) of a(subn)x^n
B(x) = sum(n=0 -> infinity) of b(subn)x^n

A(x) * B(x) = sum(n=0 -> infinity) of a(subn)x^n * sum(n=0 -> infinity) of b(subn)x^n

= sum(k=0 -> infinity) of (a(subk)*b(sub(n-k))x^n (by definition of mulitiplication)

= ?

= sum(k=0 -> infinity) of (b(subk)*a(sub(n-k))x^n

= sum(n=0 -> infinity) of b(subn)x^n * sum(n=0 -> infinity) of a(subn)x^n

= B(x) * A(x)

I can't figure out the missing step... Any advice? Thanks!

If R and R' are respectively the radius of convergence of the series



then, the radius of convergence R'' of the product series



satisfies



Now, use that both series are absolutely convergent in | x | < R'' and apply a well known theorem.