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)`

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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.