Direct Comparison Test to determine divergence or convergence

Question

the series from 1 to infinity is: 5/ (n^(1/3)-2)

The question is to determine convergence or divergence using direct comparison test. I wasn't sure how to do this because there's no value of bn that is greater than this series because the denominator is -2.

Answer 1

Before using the comparison test, let's investigate a sightly different series: Sum [ n^(1/3) ].

It is obvious that for all n, n > n - 2 and that for n>= 1, n^(1/3) > n^(1/3) - 2.

Hence,

1 / [ n^(1/3) - 2 ] > 1 / n^(1/3)

Since 1 / n^(1/3) is Divergent, then the original series is Divergent.

Answer 2

5/(n^1/3 - 2) > 1/n^1/3 for every n > 8. If you make the denominator of a fraction smaller, the result is a value larger than started with. For example start with 1/10 and reduce the denominator by 2 giving 1/8. 1/8 > 1/10.

Since ∑1/n^1/3 is a divergent p-series, the first series diverges also.