# Thread: Multiplicative comutativity of power series

1. ## 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!

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

$\displaystyle\sum_{n=0}^{+\infty}a_nx^n\;,\;\displaystyle\sum_{n=0}^{+\infty}b_nx^n$

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

$\displaystyle\sum_{n=0}^{+\infty}\left(\displaystyle\sum_{i+j=n}a_ib_jx^{i+j}\right)x^n$

satisfies

$R''\geq \min \{R,R'\}$

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