For what values of x does the series 1 + 2^x + 3^x + 4^x + ... + n^x + ... converge?

Question

A. No values of x
B. x >= -1
C. x > -1
D. x < -1
E. All values of x

I think the answer is C, because n^(-1) = 1/n, which diverges.

Answer 1

but what if x were -10

1 + 1/2^10 + 1/3^10 + 1/4^10 .....

Answer 2

Then it would converge. So the series converges for values greater than -1 that are not zero, but it also converges for all values less than -1.

Answer 3

-10 <-1, to the left on a number line