# What series should I use to compare? Direct comparison test

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.

• For A), you should indeed try to compare with $3^{-n}$, just find an expression involving it that is smaller than $3^n-1$. For B), the summand is of size $n^{-1/2}$ so can you tell whether it converges or not? Aug 4, 2019 at 14:08

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?

For A) use that $$\frac{1}{3^n-1}<\frac{2}{3^n}$$ this is equivalent to $$2<3^n$$