How do you use the integral test to determine the convergence or divergence of #1+1/sqrt2+1/sqrt3+1/sqrt4+...#?
1 Answer
May 8, 2017
The series is divergent and therefore has no finite sum.
Explanation:
This is defined by the formula
#t_n = 1/sqrt(n)#
Therefore, if the integral
#int_1^(oo) 1/sqrt(n) dn#
Has a calculable value, then the series converges. The integral can be rewritten as
#S = lim_(t->oo) int_1^t n^(-1/2) dn#
#S = [2n^(1/2)]_1^t#
#S = lim_(t->oo) 2sqrt(t) - lim_(t->oo) 2(1)^(1/2)#
The first limit obviously has value of
Hopefully this helps!