`sum_(n=2)^oo lnn/n^3` Determine the convergence or divergence of the series.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

To evaluate the series `sum_(n=2)^oo ln(n)/n^3` , 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...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

To evaluate the series `sum_(n=2)^oo ln(n)/n^3` , 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/n^2` and `a_n =ln(n)/n^3`  

It follows that `a_n < b_n`

Graph: 

Note: `f(x) =1/x^2` for red graph and `g(x)=ln(x)/x^3 ` for green graph.

Apply the p-series test where `kgt0` : the `sum_(n=k)^oo 1/n^p` is convergent if `pgt1` and divergent if `plt=1` .

For the `sum_(n=2)^oo 1/n^2` , we have the corresponding value `p=2` . It satisfies the condition `pgt1` since `2gt1` .Therefore, the series `sum_(n=2)^oo 1/n^2` converges.

Conclusion:

Because `a_n < b_n` and `sum b_n` converges, then `sum a_n = sum_(n=2)^oo lnn/n^3` converges

Approved by eNotes Editorial Team