Question

Using the integral test, how do you show whether `sum 1/(n*(n))` diverges or converges from n=1 to infinity?

Answer 1

Converges.

Explanation:

Let's first ensure `f(n)=1/(n*n)=1/n^2` is positive, decreasing, and continuous on `[1, oo).` This is the case; as `n->oo, 1/n^2` gets smaller and smaller, but never takes on a negative value. Moreover, we avoid `n=0,` so it's continuous. So, we can indeed use the Integral Test.

Let `f(x)=1/x^2`

Now, take `int_1^oo1/x^2dx=lim_(t->oo)int_1^t1/x^2dx=lim_(t->oo)(-1/x)|_1^t`

This gives

`lim_(t->oo)(-1/t+1)=1`

So, since `int_1^oo1/x^2dx` converges (has a finite value), `sum_(n=1)^oo1/n^2` also converges (but we cannot determine its value just from the performed Integral Test).

Answer 2

`sum_(n=1)^oo1/n^2` converges

Explanation:

According to the integral test, if

`int_1^oo1/x^2dx`

converges, then

`sum_(n=1)^oo1/n^2`

must also converge.

Let's compute the integral:

`int_1^oo1/x^2dx=lim_(t->oo)int_1^tx^-2dx`

`rArrlim_(t->oo)[x^-1/-1]_1^t`

`rArrlim_(t->oo)[(-1/t)-(-1/1)]`

`rArr[0-(-1)]=1`

The integral converges, so according to the integral test, the series must also converge!