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

1 Answer
Jul 18, 2018

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

Explanation:

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.