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 integern .
This means that we can be sure that:
lim_(n->oo)((n^2+3)/(2+n^2))^(n^3) >= 1
Which means that
The series fails the n-th term test, and therefore diverges.