To evaluate the series `sum_(n=1)^oo 1/(3n-2)` , we may apply **Direct
Comparison test**.

Direct Comparison test is applicable when `sum a_n` and` sum b_n ` are both positive series for all n where `a_n lt=b_n` .

If `sum b_n` converges then `sum a_n` converges.

If `sum a_n` diverges so does the `sum b_n` diverges.

Let `b_n=1/(3n-2)` and `a_n =1/(3n)` since

It follows that `a_n < b_n`

Apply `sum c* a_n = c sum a_n` , we get:

`sum_(n=1)^oo 1/(3n) =1/3sum_(n=1)^oo 1/(n)`

Apply he **p-series test**: `sum_(n=1)^oo 1/n^p` is
convergent if `pgt1` and divergent if `plt=1` .

For the `sum_(n=1)^oo 1/n ` or `sum_(n=1)^oo 1/n^1` , we have the corresponding value `p=1` . It satisfies the condition `plt=1` then the series `sum_(n=1)^oo 1/n` diverges. Therefore, the `sum_(n=1)^oo 1/(3n)` is a divergent series.

When the `sum a_n = sum_(n=1)^oo 1/(3n) ` is a divergent series then `sum b_n= sum_(n=1)^oo 1/(3n-2)` is also a divergent series.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.