# Clarification on absolute / conditional convergence test

I understand that given the absolute convergence test, if I am able to prove that the absolute of the series converges, then the series itself will converge itself as well.

What if I want to prove for conditional convergence? Is it sufficient to prove straight away that the absolute does not converge, or do I have to first prove that the series converges, and then prove that absolute does not converge, hence it must converge conditionally?

• You have to prove that the series converges and also prove that it does not converge absolutely. – Kavi Rama Murthy Oct 9 '18 at 9:06

## 2 Answers

Yes, of course you need to prove both. If you prove the series of absolute values does not converge it doesn't mean that the original series converge. For example the series $$\sum_{n=0}^\infty (-1)^n$$ simply diverges.

A series converges conditionally if both of the following conditions are true:

1. The series converges
2. The series does not converge absolutely.

To prove a series converges absolutely, you need to prove both conditions. It is not enough to prove that the series does not converge absolutely, since, for example, $$\sum_{n=1}^\infty (-1)^n n$$ does not converge absolutely, but that doesn't mean it converges conditionally.

• I have an example on hand. 1 + sin$(\frac{1}{2})$ + sin$(\pi + \frac{1}{2}) + ...$. Since sin$(\frac{1}{n})$ is simply the negative of sin$(\pi + \frac{1}{n})$, I see that the series converges to 1. In this case, am I right to say that this series converges conditionally but not absolutely? – statsguy21 Oct 9 '18 at 9:11
• @statsguy21 Did you prove that the series doesn't converge absolutely? So far, you only proved (well, sort of proved, there's still some work to do) that the series converges. – 5xum Oct 9 '18 at 9:15