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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.

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  • $\begingroup$ 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? $\endgroup$
    – kaedit
    Aug 4, 2019 at 14:08

2 Answers 2

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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?

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

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