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 thensum 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 seriessum_(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.

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