Exercise 6.5.9

Question

Review the definitions and results from Section 2.8 concerning products of series and Cauchy products in particular. At the end of Section 2.9, we mentioned the following result: If both $\sum a_{n}$ and $\sum b_{n}$ converge conditionally to $A$ and $B$ respectively, then it is possible for the Cauchy product,
$$
\sum d_{n} \quad 	ext { where } \quad d_{n}=a_{0} b_{n}+a_{1} b_{n-1}+\cdots+a_{n} b_{0}
$$
to diverge. However, if $\sum d_{n}$ does converge, then it must converge to $A B$. To prove this, set
$$
f(x)=\sum a_{n} x^{n}, \quad g(x)=\sum b_{n} x^{n}, \quad 	ext { and } \quad h(x)=\sum d_{n} x^{n} .
$$
Use Abel's Theorem and the result in Exercise $2.8.7$ to establish this result.

Ulisse Mini · Accepted Answer

By Abel's Theorem we have uniform convergence of the series defining $f$, $g$, and $h$ over the compact set $[0,1]$; therefore each of these functions is continuous and bounded over this set. We can thus conclude that for $x \in [0,1]$,
$$\lim_{N \to \infty} \sum^N_{i=0} \sum^N_{j=0} \left(a_i x^i\right) \left(b_j x^j\right) = \lim_{N \to \infty} \left(\sum^N_{n=0} a_n x^n\right) \left(\sum^N_{n=0} b_n x^n\right) = f(x) g(x)$$

Since $\sum d_n$ converges, $\lim_{n \to \infty} \left|d_{n+1}/d_n\right| \leq 1$  (otherwise $d_n$ would not be bounded). But since $\sum d_n$ only converges conditionally, $\lim_{n \to \infty} \left|d_{n+1}/d_n\right| = 1$ (if it were less than 1, then we could use the Ratio Test to prove absolute convergence). We therefore have absolute convergence of the series defining $h(x)$ for $|x| < 1$ by the Ratio Test.

From the work in Section 2.8, because we have absolute convergence, informally we have a lot of leeway in how to evaluate the double summations when $\left|x\right| < 1$. In particular,
$$\lim_{N \to \infty} \sum^N_{i=0} \sum^N_{j=0} \left(a_i x^i\right) \left(b_j x^j\right) = \sum^\infty_{i=0} \sum^\infty_{j=0} \left(a_i x^i\right) \left(b_j x^j\right) = \sum^\infty_{n=0} d_n x^n = h(x)$$

We now have the equality $f(x)g(x) = h(x)$, for $\left|x\right| < 1$. $h(x)$ is a power series, and Abel's Theorem implies $h(x)$ is continuous over $[0, 1]$. We also have continuity of $f(x)g(x)$ over $[0,1]$; thus by taking limits we have that $h(1) = f(1)g(1)$, and we thus have that $AB =\sum d_n$.

Exercise 6.5.9

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Exercise 6.5.9

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