For what value(s) of $p$ does the series $\sum_{n=1}^{\infty} \frac{n^{2p}+1}{\sqrt{n+2}}$ converge?

Question

It's a multiple choice question.

My working:

Given series is approximately $$\frac{n^{2p}}{\sqrt{n}}=\frac{1}{n^{\frac12-2p}}$$ For the series to converge, $$\frac12-2p>1\implies p<\frac14$$

But that isn't in the options. What am I doing wrong?

Answer 1

Note that

$$\sum_{n=1}^{\infty} \frac{n^{2p}+1}{\sqrt{n+2}}\ge \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+2}}$$

Answer 2

To answer the question of what you're doing wrong: Two things. Second, $1/2-2p>1$ does not imply $p>1/4$ as you claim, in fact $1/2-2p>1$ is equivalent to $2p<-1/2$, or $p<-1/4$.

But it does not follow that the series converges for $p<-1/4$. Because your first estimate, "Given series is approximately..." is only correct for $p>0$; in fact if $p<0$ then the numerator $n^{2p}+1$ is approximately $1$.

You have to be careful with those inequality things...