Question

How do you use the integral test to determine the convergence or divergence of `Sigma 1/n^3` from `[1,oo)`?

Answer 1

The series:

`sum_(n=1)^oo 1/n^3`

is convergent.

Explanation:

Given the series:

`sum_(n=1)^oo 1/n^3`

we can choose as test function:

`f(x) = 1/x^3`

We have that `f(x)` is positive and decreasing in `(1,+oo)` that:

`lim_(x->oo) 1/x^3 = 0`

and finally that `f(n) = a_n`. Thus all the hypotheses of the integral test theorem are satisfied, and the convergence of the series is proved to be equivalent to the convergence of the improper integral:

`int_1^oo (dx)/x^3`

Calculating the integral we find:

`int_1^oo (dx)/x^3 = [-1/(2x^2) ]_1^oo`

`int_1^oo (dx)/x^3 = lim_(x->oo) (-1/(2x^2)) +1/2 = 1/2`

Thus the integral is convergent and then the series is proven to be convergent as well.