Find the Cauchy product of the series sum from n=0
Find the Cauchy product of the series $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+1}}$ with itself, and show that the resulting series does not converge although the original series converges.
Explanation of Method
The original series is conditionally (not absolutely) convergent by Leibniz's test, since the comparison series $\sum1/\sqrt{n+1}$ is a divergent $p$-series with $p=\frac12$. Forming the Cauchy product term $c_n$ and bounding each summand below via AM-GM shows $|c_n|$ grows like $\sqrt n$ rather than shrinking to zero — so the product series fails the divergence test outright, a striking illustration that absolute convergence of at least one factor (Mertens' theorem) cannot be dropped.
- Verify the original series converges conditionally, not absolutely.
- Form the Cauchy product term $c_n$ and bound $|c_n|$ below via AM-GM.
- Conclude divergence via the $n$-th term test.
Step 1 · The original series converges conditionally
With $a_n=1/\sqrt{n+1}$: positive, decreasing, $\to0$, so Leibniz's test gives convergence of $\sum(-1)^n/\sqrt{n+1}$. But $\sum1/\sqrt{n+1}$ is a $p$-series with $p=\frac12<1$, so it diverges — the convergence is only conditional.
Step 2 · Form and bound the Cauchy product
$$ c_n=\sum_{k=0}^na_kb_{n-k}=(-1)^n\sum_{k=0}^n\frac1{\sqrt{(k+1)(n-k+1)}}. $$ By AM-GM, $(k+1)(n-k+1)\le\left(\frac{n+2}2\right)^2$, so each term is $\ge\dfrac2{n+2}$, giving $|c_n|\ge(n+1)\cdot\dfrac2{n+2}$; a finer estimate (via a Beta-function/integral comparison) shows $|c_n|\sim\pi\sqrt n$.
Step 3 · Divergence test
Since $|c_n|\sim\pi\sqrt n\to\infty$, we have $c_n\not\to0$, so by the $n$-th term test, $\sum c_n$ diverges — despite $\sum(-1)^n/\sqrt{n+1}$ itself converging.
Summary Table
| Series | Behavior |
|---|---|
| $\sum(-1)^n/\sqrt{n+1}$ | Converges conditionally |
| Cauchy product $\sum c_n$ | Diverges ($|c_n|\sim\pi\sqrt n$) |
Additional Notes
- This is the classical Cauchy counterexample referenced in Problem 137's discussion of Mertens' theorem: it is the sharpest and most well-known instance of two conditionally convergent series whose Cauchy product fails to converge.
- The exponent $\frac12$ is critical here exactly as noted in Problem 137: it is the boundary case for conditional convergence of $\sum(-1)^n/n^p$, and simultaneously the unique exponent at which the AM-GM bound on the product terms fails to vanish.
- Mertens' theorem (used positively in Problem 105 and referenced in Problem 137) requires absolute convergence of at least one of the two factors; this problem demonstrates concretely why that hypothesis cannot simply be weakened to conditional convergence of both factors.
Result
The Cauchy product diverges despite the original series converging: