Question

Determine whether the series is convergent or divergent.

Answer 1

\[\sum_{1}^{\infty} \frac{ n-2 }{ \sqrt[3]{3n^7 -1} }\]

Answer 2

I believe I'm supposed to use the comparison test.

Answer 3

Use the comparison test. Just using my knowledge of it, it is going to be convergent. But use the comparison test as you believed. "Never doubt the thinkers mind."

Answer 4

Considering the degrees of the numerator and denominator, you should compare to
\[\sum_{n=1}^\infty \frac{n^1}{n^{7/3}}=\sum_{n=1}^\infty \frac{1}{n^{4/3}}\]

Answer 5

I also believe that since you are comparing to being always bigger than 1/n, that is is convergent. That's what I was basically taught, except with more to it obviously.

Answer 6

Correct, we'd have to show that
\[\frac{n-2}{\sqrt[3]{3n^7-1}}>\frac{1}{n^{4/3}}\]
Since the series with the right hand sequence converges, showing the above means the left hand sequence converges too.

Answer 7

Brings back the memories of high school calculus 2.