Mertens Theorem

If (i) $\sum a_n$ converges absolutely (ii)$\sum a_n = A$ (iii)$\sum b_n = B$ (iv)$c_n = \sum_{k=0}^n a_k b_{n-k}$, Then $\sum c_n = AB$.

Here, what if (iii) is changed to ' $\sum b_n$ diverges'? If it is inconclusive, please give me an example.

Plus, if $\sum a_n$ and $\sum b_n$ both diverge, does $\sum c_n$ diverge?

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Observe that $(c_n)$ is the Cauchy product (or discrete convolution) of $(a_n)$ and $(b_n)$. That is, it satisfies (at least as a formal identity)

$$\sum c_n x^n = \left( \sum a_n x^n \right)\left( \sum b_n x^n \right). \tag{1}$$

This shows that it is natural to consider power series to test if your new condition

(iii') $\sum b_n$ diverges

• First, let us consider the following example: $$\sum a_n x^n = 1 - x, \qquad \sum b_n x^n = \frac{1}{1-x}.$$ Then $a_n = (1, -1, 0, 0, \cdots)$ and $b_n = (1, 1, 1, 1, \cdots )$ satisfies (i), (ii) and (iii'). Now, $(c_n)$ defined by (iv) satisfies $$\sum c_n x^n = \left( \sum a_n x^n \right)\left( \sum b_n x^n \right) = (1+x)\left(\frac{1}{1+x}\right) = 1,$$ so that $c_n = (1, 0, 0, 0, \cdots)$. Therefore in this case, $\sum c_n$ converges.
• However, if we change $(b_n)$ so that $$\sum b_n = \frac{1}{(1-x)^2},$$ then we have $(b_n) = (1, 2, 3, 4, \cdots)$ instead, which again satsifies (iii'), but now the application of $(1)$ again gives $c_n = (1, 1, 1, 1, \cdots)$. Therefore $\sum c_n$ diverges.
• Finally, consider the following example \begin{align*} \sum a_n x^n &= \frac{1-x}{1+x} = 1 - 2x + 2x^2 - 2x^3 + \cdots, \\ \sum b_n x^n &= \frac{1+x}{1-x} = 1 + 2x + 2x^2 + 2x^3 + \cdots. \end{align*} Then both $\sum a_n$ and $\sum b_n$ diverge, but $\sum c_n$ converges since from $(1)$, $$\sum c_n x^n = \left( \sum a_n x^n \right)\left( \sum b_n x^n \right) = 1,$$ so that $c_n = (1, 0, 0, 0, \cdots)$.