Question

How do you know if the summation `(-12)^n/n` where n is between 3 to infinity is convergent or divergent?

Answer 1

`sum_(n=3)^oo(-12)^n/n` is divergent.

You can determine that this series diverges because the terms do not go to `0` as `n` increases without bound. That is: `lim_(nrarroo)((-12)^n)/n!=0`.