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`
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.