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.
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