How do you use the integral test to determine the convergence or divergence of #Sigma 1/sqrtn# from #[1,oo)#?

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Answer 1 · 2018-07-18T08:48:15

The series #sum_(n=1)^oo 1/sqrtn# is divergent.

Let #f(x) = 1/sqrtx#.

We have that:

1) #f(x) >0#

2)# lim_(x->oo) f(x) = 0#

3) # (df)/dx = -1/(2xsqrtx) < 0# so that #f(x)# is strictly decreasing in #[1,+oo)#

4) # f(n) = 1/sqrtn#

Under these conditions the convergence of the series:

#sum_(n=1)^oo 1/sqrtn#

is equivalent to the convergence of the improper integral:

#int_1^oo dx/sqrtx#

Evaluate the indefinite integral:

#int dx/sqrtx = 2sqrtx#

and we can see that the integral:

#int_1^oo dx/sqrtx = -2 + 2lim_(x->oo) sqrtx#

is not convergent.

Then also the series #sum_(n=1)^oo 1/sqrtn# is not convergent, and as #1/sqrtn >0# we can conclude that it is divergent.