
Mark M. answered 10/27/20
Mathematics Teacher - NCLB Highly Qualified
(3 + 5n) / 5n
(3 / 5n) + (5n / 5n)
(3 / 5n) + 1
While the first term converges to 0 the summation of the second term is an unlimited number of 1's.
i.e.: [3/5 + 1] + [3/25 + 1] + [3/125 + 1], [3/625 + 1].......
Diverges
Giulia T.
thanks10/27/20