Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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

leads to an inconclusive statement.

  • 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 $c_n = (1, 0, 0, 0, \cdots)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.