Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$a_n, b_n$ - sequences
Suppose $a_n+b_n$ converges. Does $a_n b_n$ converge also?

I tried thinking if I can learn something about $a_n$ and $b_n$ by the assumption $a_n b_n$ converges.
I also tried to develop this equation $|a_n b_n - L| < \epsilon$ assuming it is converging.

I didn't get any bright conclusions.
Will be glad help.

share|cite|improve this question
What does the first "sentence" mean? Certainly those are sequences, not sets. – Marc van Leeuwen Nov 25 '13 at 13:51
up vote 20 down vote accepted

$a_n =n, b_n =-n$

$a_n+b_n=0$ but then...

$a_nb_n=-n^2\rightarrow -\infty$

share|cite|improve this answer
When we are talking about convergence, I think it needs to converges to a real number. A sequences that tends to infinity does not converges – Giiovanna Nov 25 '13 at 13:34
Yes, indeed. This only shows an example when the product does not converges. But it is really easy to show one that the product does converges. Just take 2 constant sequences. The sum is a constant and so does the product. Then, as we can see, the product can or not converge – Giiovanna Nov 25 '13 at 13:39
Could you please let me know what are you expecting when you say "general way" – Praphulla Koushik Nov 25 '13 at 14:04
@Giiovanna: The question is if $a_n+b_n$, does $a_nb_n$ also converge? Certainly, anyone can come up with cases where both converge. However, to negate a false implication, one needs to satisfy the hypotheses (in this case, $a_n+b_n$ converges), yet show that the conclusion is not necessarily true (in this case, $a_nb_n$ does not converge). Finding a pair of sequences for which $a_nb_n$ converges does not apply to the question. – robjohn Nov 25 '13 at 15:34
@Giiovanna If you were asked to prove that 2x is not equal to x^2 how would you do that? Then what would you say to the person that says well it's true for x=2! – Cruncher Nov 25 '13 at 18:01

As shown in Praphulla Koushik's answer $$ a_n=n, b_n=-n $$ cancellation is a problem.

However, even if you restrict $a_n,b_n\ge0$ then the answer is no. For example, if $$ a_n=2+(-1)^n,b_n=3-(-1)^n $$ then $a_n+b_n=5$ yet $a_nb_n$ oscillates between $4$ and $6$.

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.