1. ## Converge or diverge?

Does the series from 1 to infinity 1/(5n^2+1)^(1/3) converge or diverge?

I thought by using the divergence test and checking if the limit goes to 0 as n approaches infinity that the series converges. But the answer is "diverges." What test could I use to get divergence for this series?

2. Originally Posted by jlmills5
Does the series from 1 to infinity 1/(5n^2+1)^(1/3) converge or diverge?

I thought by using the divergence test and checking if the limit goes to 0 as n approaches infinity that the series converges. But the answer is "diverges." What test could I use to get divergence for this series?

$\frac{1}{(5n^2+1)^{1\slash 3}}\geq \frac{1}{(10n^2)^{1\slash 3}}$ $=\frac{1}{\sqrt[3]{10}}\,\frac{1}{n^{2\slash 3}}$ , and as $\sum\limits_{n=1}^{\infty}\frac{1}{n^p}$ converges iff $p>1$ , our series diverges by the comparison test.

Tonio

3. Thank you! I have such issues trying to find a good comparison series when using direct and limit comparison.

4. Hello, jlmills5!

Converge or diverge? . $S \;=\;\sum^{\infty}_{n=1} \frac{1}{(5n^2+1)^{\frac{1}{3}}}$
I forced it into a comparison test . . .

For $n \geq 1\!:\;\;1 \:<\:3n^2$

Add $5n^2$ to both sides: . $5n^2 + 1 \:< \:8n^2$

Take the cube root of both sides: . $\left(5n^2+1\right)^{\frac{1}{3}} \:< \:\left(8n^2\right)^{\frac{1}{3}}$

. . That is: . $(5n^2+1)^{\frac{1}{3}} \;<\;2n^{\frac{2}{3}}$

Take the reciprocal of both sides: . $\frac{1}{\left(5n^2+1\right)^{\frac{1}{3}}} \;{\color{red}>} \;\frac{1}{2n^{\frac{2}{3}}}$

Summate both sides: . $\sum^{\infty}_{n=1}\frac{1}{\left(5n^2 +1\right)^{\frac{1}{3}}} \;>\;\sum^{\infty}_{n=1}\frac{1}{2n^{\frac{2}{3}}}$

We have: . $S \;>\;\frac{1}{2}\sum^{\infty}_{n=1}\frac{1}{n^{\fr ac{2}{3}}}$ . . . a divergent $p$-series

Since $S$ is greater than a divergent series, $S$ diverges.

5. Okay, I'm looking over this again and I'm not sure I understand how this means divergence.
The original series is greater than or equal to the comparison series. I thought the comparison test only worked if the comparison series was greater than the original. And only worked then if the comparison series converges (and hence the original converges.) This is what I was taught in class anyway.

6. Originally Posted by jlmills5
Okay, I'm looking over this again and I'm not sure I understand how this means divergence.
The original series is greater than or equal to the comparison series. I thought the comparison test only worked if the comparison series was greater than the original. And only worked then if the comparison series converges (and hence the original converges.) This is what I was taught in class anyway.
Since the "smaller" series diverges, then the "larger" series must diverge as well.

Similarly, if you find a "larger" series that converges, then the "smaller" series must converge as well.

7. Thanks! I think I've been looking at this too long. I understood and first and then second guessed myself. I got it now!