Dose sum ((n^2+3)/(2+n^2))^(n^3) with n = 0 -> infinity converge ?

1 Answer
Jun 5, 2017

No.

Explanation:

Use the n-th term test:

For any integer n, (n^2+3)/(2+n^2) will always be greater than 1.

Therefore, ((n^2+3)/(2+n^2))^(n^3) will always be greater than 1 for any positive integer n.

This means that we can be sure that:

lim_(n->oo)((n^2+3)/(2+n^2))^(n^3) >= 1

Which means that lim_(n->oo)((n^2+3)/(2+n^2))^(n^3) != 0 for sure.

The series fails the n-th term test, and therefore diverges.