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.