Question

How do you use the integral test to determine the convergence or divergence of `Sigma 1/n^2` from `[1,oo)`?

Answer 1

The series `sum_(n=1)^oo 1/(n^2)` is convergent.

Explanation:

The integral test states that, given:

`sum_(n=1)^oo a_n`

if we have a function `f(x)` such that:

(i).... `f(x)` is positive
(ii)... `f(x)` is decreasing
(iii) `lim_(x->+oo) f(x) = 0`
(iv)...`f(n) = a_n`

then a necessary and sufficient condition for the series above to converge is that the integral:

`int_1^(+oo) f(x)dx`

converges as well.

It's easy to see that for the series:

`sum_(n=1)^oo 1/(n^2)`

the function `f(x) = 1/(x^2)` satisfies all the conditions above, and we have that:

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

so that the series must be convergent as well.