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

1 Answer
Dec 22, 2016

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.