Suppose $\sum a_n$ converges absolutely and $\sum b_n$ converges. Give an example where the Cauchy product does not converge absolutely.

Solution 1:

Take $a_n = (-1)^n/n^2$ and $b_n = (-1)^n/n$. Then $\sum a_n$ converges absolutely and $\sum b_n$ converges conditionally.

The Cauchy product

$$\sum_{n=1}^\infty \sum_{k=1}^n \frac{(-1)^k}{k^2} \frac{(-1)^{n+1-k}}{n+1-k} = \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^n \frac{1}{k^2(n+1-k)}, $$

converges by the AST since the terms $\sum_{k=1}^n 1/(k^2(n+1-k))$ are decreasing.

We also know the Cauchy product converges by the general theorem guaranteeing convergence if one series is absolutely convergent and the other is convergent.

However, the Cauchy product is not absolutely convergent since

$$\sum_{n=1}^\infty\sum_{k=1}^n \frac{1}{k^2(n+1-k)} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{ 1}{3} + \frac{1}{8} + \frac{1}{9} + \ldots > 1 + \frac{1}{2}+ \frac{1}{3} + \ldots$$

diverges.

Solution 2:

Hint: Consider the Cauchy product of $1+0+0+0+\dots$ and $1-\frac12+\frac13-\frac14+\dots$.

We can take this even further: $$ \overbrace{\sum_{k=0}^\infty\frac{1+(-1)^k}{2^{k+1}}}^{\substack{\text{converges}\\\text{absolutely}}}\,\,\overbrace{\sum_{j=0}^\infty\frac{(-1)^j}{j+1}}^{\substack{\text{converges}\\\text{conditionally}}} $$ The Cauchy product is $$ \sum_{j=0}^\infty(-1)^j\overbrace{\sum_{k=0}^{\lfloor j/2\rfloor}\frac1{j-2k+1}\frac1{2^{2k}}}^{a_j\ge\frac1{j+1}} $$ where, for $j\ge5$, $$ \begin{align} a_{j-1}-a_j &=\sum_{k=0}^{\lfloor(j-1)/2\rfloor}\frac1{j-2k}\frac1{2^{2k}}-\sum_{k=0}^{\lfloor j/2\rfloor}\frac1{j-2k+1}\frac1{2^{2k}}\\ &=\sum_{k=0}^{\lfloor(j-1)/2\rfloor}\frac1{(j-2k)(j-2k+1)}\frac1{2^{2k}}-\frac{1+(-1)^j}{2^{j+1}}\\ &\ge\frac1{j(j+1)}-\frac1{2^j}\\[12pt] &\ge0 \end{align} $$ Thus, the Cauchy product converges by the Dirichlet Test, but does not converge absolutely since $a_j\ge\frac1{j+1}$