Question

What is the interval of convergence of `sum_1^oo (3n)/n^(x)`?

Answer 1

`x>2`

Explanation:

For `x-1>0` which is the condition for which `k^{-(x-1)}` decreases monotonically, we have `sum_1^n 3 k^{-(x-1)} le int_1^n 3k^{-(x-1)}dk +3= n^{2-x}/(2-x)-1/(2-x)+3` and also for `2-x < 0` we have

`lim_{n->oo}n^{2-x}/(2-x)-1/(2-x)+3 = 1/(x-2)+3`. So
we have

`sum_1^n 3 k^{-(x-1)} le 3/(x-2)+3`