Question

HELP! If the series is positive term, determine whether it is convergent or divergent. If the series contains negative terms, determine whether it is absolutely convergent, conditionally convergent or divergent. State what test you used to make that determination.

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 1 }{ \sqrt[3]{n(n+1)(n+2)} }\]

Answer 2

Compare this series to a known convergent/divergent series. I'd try \(\sum \dfrac{1}{n}\).

Answer 3

\(n(n+1)(n+2) > n \implies \Large \frac{1}{n(n+1)(n+2)}< \frac{1}{n} \)
problem is here that the harmonic series diverges, however this would not be a statement about the smaller sum.

Answer 4

But @SithsAndGiggles suggest the correct way, this is usually done by comparison with a minor/major expression.

Answer 5

I first tried Ratio Test, but it resulted to be inconclusive.

Answer 6

So you need an inequality in the different direction :-)

Answer 7

Good, since you've already tried the Ratio test, and you figured out that it is inclusive, you automatically know that it is no use applying the root test (they are basically equal). So you want to tackle the comparison test.

Answer 8

Do you mind showing me how to use the Comparison Test in this particular problem?

Answer 9

Have you considered the asymptotic behavior of your expression? Maybe that is what @SithsAndGiggles meant to suggest, I apologize if that was the case. \[\Large \lim_{n \to \infty} \frac{n(n+1)(n+2)}{n^3} =1 \implies a_n \sim b_n \]
In which you can say that \(a_n\) and \(b_n\) have the same behavior when it comes to their convergence/divergence. The rest would be analogous to as above @SithsAndGiggles suggested.

Answer 10

I actually did have the direct comparison test in mind, but soon enough came to the same conclusion, that it wouldn't work. The limit comp test should work, though.

Answer 11

It's what I first tried myself but I couldn't find a satisfying substitution to complete the task without using the comparison of the asymptotic behavior :(

Answer 12

Thank you for the help, guys! I will try to figure it out! Thanks! :-)