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