It is well-known that
the Cauchy product of two absolutely convergent series is absolutely convergent.
However, my professor added (without giving a proof) that if the series are convergent (but not absolutely convergent), then the Cauchy product is either convergent or irregular (i.e. indeterminate), but not divergent to infinity.
How do you prove this statement?
We can use Abel's theorem for the proof. Suppose the two series $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ are condtionally but not absolutely convergent. For $\lvert x\rvert < 1$, we define
$$f(x) = \sum_{n=0}^\infty a_n x^n\quad \text{and}\quad g(x) = \sum_{n=0}^\infty b_n x^n.$$
By Abel's theorem, we have
$$\lim_{x\to 1^-} f(x) = \underbrace{\sum_{n=0}^\infty a_n}_A\quad\text{and}\quad \lim_{x\to 1^-} g(x) = \underbrace{\sum_{n=0}^\infty b_n}_B,$$
and therefore, with $h(x) = f(x)\cdot g(x)$,
$$\lim_{x\to 1^-} h(x) = A\cdot B.$$
The coefficients in the Power series expansion of $h$ are the terms of the Cauchy product of the two series,
$$h(x) = \sum_{n=0}^\infty c_n x^n, ~ \text{ where } c_n = \sum_{k=0}^n a_k b_{n-k}.$$
We denote the partial sums of $\sum c_n$ by $s_n$, and define
$$H(x) = \sum_{n=0}^\infty s_n x^n.$$
Assuming that we had $s_n \to +\infty$, for every $M \in \mathbb{R}^+$, there is an $N_M$ such that $s_n \geqslant M$ for all $n \geqslant N_M$. Then for $0 < x < 1$ we have
$$H(x) = \sum_{n < N_M} s_nx^n + \sum_{n=N_M}^\infty s_n x^n \geqslant \sum_{n < N_M} s_n x^n + \frac{x^{N_M}\cdot M}{1-x},$$
and hence
$$\liminf_{x\to 1^-} (1-x)H(x) \geqslant M.$$
But - setting $s_{-1} = 0$ - we have
$$h(x) = \sum_{n=0}^\infty c_n x^n = \sum_{n=0}^\infty (s_n -s_{n-1}) x^n = \sum_{n=0}^\infty s_n x^n - \sum_{n=1}^\infty s_{n-1}x^n = (1-x)H(x),$$
and so the assumption $s_n \to +\infty$ contradicts the fact that
$$\lim_{x\to 1^-} h(x) = A\cdot B.$$
