Explain why the Comparison Test can or cannot be used to decide if the series converges or diverges. You do not have to determine if the series converges or diverges.

$($a$){\displaystyle \underset{n=1}{\overset{}{}}}\frac{(-1{)}^n}{n^2}$

$($b$){\displaystyle \underset{k=1}{\overset{}{}}}\frac{ln(k)}{k}$

$($c$){\displaystyle \underset{n=1}{\overset{}{}}}sin(n)$