# Convergence tests

• May 15th 2011, 02:09 PM
Conn
Convergence tests
When I'm testing $\sum_{n = 0}^\infty \cos \frac{1}{n}$ for convergence, how come I can apply the nth term test and show that it is divergent, i.e.

$\lim_{n \to \infty} \cos \frac{1}{n} = \cos 0 = 1,$

but when I'm testing $\sin \frac{1}{n}$, I can't apply the same reasoning, i.e.

$\lim_{n \to \infty} \sin \frac{1}{n} = \sin 0 = 0 \Rightarrow \sin \frac{1}{n} converges,$

but instead have to use the limit comparison test and using L'Hopital's Rule, ending up with an implication that $\sin \frac{1}{n}$ is divergent
• May 15th 2011, 02:22 PM
Jose27
Do you know the difference between necessary and sufficient conditions?
• May 15th 2011, 02:22 PM
Mondreus
You can only show divergence that way. $\lim_{n\to \infty}a_n=0$ is necessary for convergence, but convergence does not follow from it.