`sum_(n=1)^oo ln((n+1)/n)` Determine the convergence or divergence of the series.

To determine if the series `sum_(n=1)^oo ln((n+1)/n)` converges or diverges, we may apply the 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.

For the given series `sum_(n=1)^oo ln((n+1)/n)` , we let `b_n= ln((n+1)/n)` .

  Let `a_n= ln(1/n)` since  `ln(1/n) lt= ln((n+1)/n)`...

To determine if the series `sum_(n=1)^oo ln((n+1)/n)` converges or diverges, we may apply the 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.

For the given series `sum_(n=1)^oo ln((n+1)/n)` , we let `b_n= ln((n+1)/n)` .

  Let `a_n= ln(1/n)` since  `ln(1/n) lt= ln((n+1)/n)` .

To evaluate if the series `sum_(n=1)^oo ln(1/n)` converges or diverges, we may apply Divergence test:

`lim_(n-gtoo) a_n !=0` or does not exist then the series` sum a_n` diverges 

We set-up the limit as:

`lim_(n-gtoo)ln(1/n) =lim_(n-gtoo)ln(n^(-1))`

                         ` = (-1)lim_(n-gtoo) ln(n)`

                         ` = -oo`

With the limit value `L =-oo` , it satisfy `lim_(n-gtoo) a_n !=0` .``

Thus, the series `sum_(n=1)^oo ln(1/n)` diverges      

Conclusion based from Direct Comparison test:

The series`sum_(n=1)^oo a_n = sum_(n=1)^oo ln(1/n)`  diverges then it follows that `sum_(n=1)^oo b_n =sum_(n=1)^oo ln((n+1)/n)` also diverges.