Use direct comparison test to prove if the following series converge or not.
A) $\sum_{n=0}^\infty \frac{1}{3^n -1}$
B) $\sum_{n=0}^\infty\frac{1}{\sqrt{n+2}}$
In A) I wrote $3^n -1<3^n$ so $\frac{1}{3^n -1}>\frac{1}{3^n}$, but that is useless because $\frac{1}{3^n}$ converges and it's smaller than $\frac{1}{3^n-1}$ so I can't conclude anything.
And then in B) I don't know what series I should use to compare.
You have the right idea for A). Try to compare it with a geometric series. How about using/proving the inequality $2^n \leq 3^n - 1$?
Let me give a hint for B):
We have $$\frac{1}{\sqrt{n+2}} \geq \frac{1}{\sqrt{n+n}} = \frac{1}{\sqrt{2}}\frac{1}{\sqrt{n}}$$ for all $n \geq 2$. Does this help you?