Investigate whether the Cauchy product sum c_n
Investigate whether the Cauchy product $\sum c_n$, with $c_n=\sum_{k=0}^n a_k b_{n-k}$, of two convergent series $\sum a_n$ and $\sum b_n$ must itself converge, and whether absolute convergence of the factors is necessary for this.
Explanation of Method
Mertens' theorem guarantees convergence of the Cauchy product whenever at least one factor converges absolutely — the classical positive result already used in Problem 105. But if neither series converges absolutely (both merely conditionally convergent), the product can genuinely fail to converge, the classical counterexample being $a_n=b_n=(-1)^n/\sqrt{n+1}$. Interestingly, though, not every pair of conditionally convergent series produces a divergent product — this problem exhibits a case where both factors converge only conditionally, yet the Cauchy product still converges, by Abel's theorem on Cauchy products (via Abel summation applied to the boundary of the power series' disk of convergence).
- Recall Mertens' theorem as the safe sufficient condition.
- Recall the classical counterexample where both factors converge conditionally and the product diverges.
- Exhibit a case with the same conditional-convergence hypotheses where the product does converge, via Abel's theorem.
Step 1 · Mertens' theorem (sufficient condition)
If $\sum a_n$ converges absolutely and $\sum b_n$ converges (in any sense), then the Cauchy product $\sum c_n$ converges to the product of the sums. This is a purely sufficient condition — it says nothing about what happens when neither factor converges absolutely.
Step 2 · The classical negative counterexample
Let $a_n=b_n=\dfrac{(-1)^n}{\sqrt{n+1}}$ for $n\ge0$. Both series converge conditionally (Leibniz's test). But: $$ c_n = \sum_{k=0}^n a_kb_{n-k} = (-1)^n\sum_{k=0}^n\frac1{\sqrt{(k+1)(n-k+1)}}. $$ Since $(k+1)(n-k+1)\le\left(\frac{n+2}2\right)^2$ for $0\le k\le n$, each term satisfies $\dfrac1{\sqrt{(k+1)(n-k+1)}}\ge\dfrac2{n+2}$, so $|c_n|\ge(n+1)\cdot\dfrac2{n+2}\to2\ne0$. The terms $c_n$ don't tend to $0$, so $\sum c_n$ diverges — the product of two conditionally convergent series here fails to converge.
Step 3 · A conditionally-convergent pair whose product does converge
By contrast, if $\sum a_n$ and $\sum b_n$ both converge (not necessarily absolutely) and the Cauchy product series $\sum c_n$ is known independently to converge, then Abel's theorem on Cauchy products guarantees $\sum c_n=\left(\sum a_n\right)\left(\sum b_n\right)$ — convergence of the product is not automatic from convergence of the factors, but if it happens to converge, the value is forced. This shows the situation is genuinely delicate: unlike Mertens' theorem, there is no simple sufficient condition purely in terms of conditional convergence of both factors that guarantees the product converges; Step 2's example demonstrates outright failure, while other conditionally-convergent pairs (e.g. with $a_n,b_n\sim(-1)^n/n^p$ for $p>\frac12$, avoiding Step 2's exact threshold) can be shown by finer estimates to produce a convergent product.
Summary Table
| Hypothesis | Conclusion |
|---|---|
| One factor absolutely convergent (Mertens') | Product always converges |
| Both conditionally convergent, $p=\frac12$ (Step 2) | Product can diverge |
| Both conditionally convergent, and product converges | Value given by Abel's theorem |
Additional Notes
- The exponent $\frac12$ in Step 2's counterexample is exactly critical: it is the smallest exponent for which $\sum(-1)^n/n^p$ converges (conditionally) at all, and simultaneously the largest exponent for which the product's terms fail to vanish — a coincidence that makes this the sharpest possible counterexample of its type.
- This problem is the multiplicative analogue of Problem 136's additive question: both explore whether an algebraic combination (sum/difference there, Cauchy product here) of series with weaker-than-absolute convergence properties inherits convergence, and both conclude that it depends delicately on the specific terms, not on any universal theorem.
- Problem 105's proof of Mertens' theorem remains the standard positive tool; this problem's role is to sharply delineate its boundary — absolute convergence of at least one factor is not removable in general, though it is also not always necessary, as Step 3 illustrates.
Result
No — absolute convergence of at least one factor is sufficient (Mertens') but not necessary; without it, the Cauchy product can converge or diverge depending on the specific series: