Proof that the product of two Maclaurin series forms a valid Maclaurin Series

Question

Given two Maclaurin Series,

$$ f_0(x):= \sum_{k=0}^{\infty} a_{k} x^k;\quad \forall\ x\in [u_0, v_0], \quad f_1(x):= \sum_{k=0}^{\infty} b_{k} x^k;\quad \forall\ x\in [u_1, v_1]. $$

I have, many times, used that, if $u_0=u_1$ and $v_0=v_1,$ then:

$$ f_0(x) + f_1(x) = \sum_{k=0}^{\infty} (a_{k} + b_k) x^k\quad \forall\ x\in I:= [u_0, v_0] \cap [u_1, v_1] $$

This is clearly true, because it is true for all convergent series, not just Maclaurin series.

$$$$

However, I have also used many times, without proof, the following:

$$(1)\qquad f_0(x) f_1(x) = \sum_{k=0}^{\infty} c_k x^k\quad \forall\ x\in I $$

where

$$ c_k = \sum_{j = 0}^{k} a_{k-j} b_j. $$

For example, in an exam question to find the Maclaurin expansion of $\ e^x\sin x,\ $ up until the $x^5$ term, I think it is standard to do the following:

$$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots $$

and

$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots. $$

"Therefore",

$$ e^x \sin x = \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots \right) \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \right) $$

$$ = x + x^2 + \left( \frac{1}{2!} - \frac{1}{3!} \right) x^3 + \left( \frac{1}{3!} - \frac{1}{3!} \right) x^4 + \left( \frac{1}{5!} - \frac{1}{2!} \frac{1}{3!} + \frac{1}{4!} \right) x^5 + \ldots. $$

However, it's not super obvious to me that, $(a)$ in general, this expression (i.e. that the RHS of $(1)$ ) converges; $(b)$ that $(1)$ is true.

I cannot think of a proof of either $(a)$ or $(b)$ off the top of my head. Are these always true? If so, how do you prove them?

Answer 1

According to the Wikipedia page for Cauchy product, it suffices for one of the series to converge absolutely. However, if both series converge conditionally, it is possible that the Cauchy product series does not converge.

Answer 2

Hint. The fact that formally multiplying power series does the right thing on the open interval determined by the minimum of their radii of convergence follows from the fact that you can differentiate a power series term by term.

Answer 3

I couldn't prove it, but I did find an approach that gets close. There's a connection you can make with the limit multiplication law, and partial sums. If $P_0(n)$ is the nth partial sum of the maclaurin series of $f_0$, and $P_1(n)$ is the nth partial sum of the maclaurin series of $f_1$, then we can rephrase everything in terms of limits. Given $\lim_{n\to\infty} P_0(n) = f_0$ and $\lim_{n\to\infty} P_1(n) = f_1$ then $\lim_{n\to\infty}{P_0(n) \cdot P_1(n)} = f_0 \cdot f_1$ by the limit laws, so if you can prove that some partial sum of the taylor series for $e^x sin(x)$ above is equal to a partial sum of $e^x$ times a partial sum of $sin(x)$, that would prove it true.

Table	$x^0$	$x^1$	$x^2$	$x^3$
$x^0$	$x^0$	$x^1$	$x^2$	$x^3$
$x^1$	$x^1$	$x^2$	$x^3$	$x^4$
$x^2$	$x^2$	$x^3$	$x^4$	$x^5$
$x^3$	$x^3$	$x^4$	$x^5$	$x^6$

Unfortunately, this does not work. Let's call the taylor series for $e^x sin(x)$ $C(n)$, and the partial sums $P_0(n)P_1(n)$ $Q(n)$. Each term of $C(n)$, $C(n+1) - C(n)$, is a diagonal on the above table, adding the terms of the same power, resulting in a triangle partial sum. However, each partial sum of $Q(n)$ is a square, because it's the cross section of $P_0(n)$ and $P_1(n)$. That means each "term" of $Q(n)$ is an edge of a square. $Q(n)$ has more than one power of x in each term, making it not a taylor series. So $C(n) \neq Q(n)$.

This doesn't work, but I think maybe you could prove $\lim_{n\to\infty} C(n) = \lim_{n\to\infty} Q(n)$ instead? I see a connection to absolute/conditional convergence with different orders of adding things.