Question

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

Answer 1

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.