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

Recall that the Divergence test follows the condition:


If `lim_(n-gtoo)a_n!=0` then sum `a_n` diverges.


For the given series `sum_(n=1)^oo ln(1/n)` , we have `a_n = ln(1/n)` .


To evaluate it further, we may apply Law of exponent: `1/x^n = x^-n` .


`a_n = ln(1/n)` is the same as  `a_n = ln(1/n^1)`


Then, `a_n = ln(n^(-1))` .


 Apply natural logarithm property: `ln (x^n) = n *ln(x)` .


`a_n = (-1) *ln(n)`


or `a_n = -ln(n)`


Applying the divergence test, we determine the limit of the series as:


`lim_(n-gtoo) [ -ln(n)] = -lim_(n-gtoo) ln(n)`


                            `= - oo`


Conclusion:


The limit value (L) being `-oo` implies that the series `sum_(n=1)^oo ln(1/n)` is  divergent.


 We can also verify with the graph of `f(n) = ln(1/n)` :



As the "`n` " values increase, the function value decreases to negative infinity and does not approach any finite value of L.

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