# 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

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.
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