Question

How do you use the integral test to determine the convergence or divergence of `1+1/4+1/9+1/16+1/25+...`?

Answer 1

Our first step is to write the series in series notation. Notice that:

`1+1/4+1/9+1/16+1/25+...`

`=1/1^2+1/2^2+1/3^2+1/4^2+1/5^2+...`

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

So, we want to determine the convergence or divergence of the series `sum_(n=1)^oo1/n^2` using the integral test.

The integral test requires some conditions: for `sum_(n=N)^oof(n)`:

• `f(n)` has to be decreasing on `[N,oo)`
• `f(n)>=0` on the same interval
• `f(n)` is continuous on the interval

For `f(n)` we see that these are all true so the integral test will apply. The actual integral test states that:

The series `sum_(n=N)^oof(n)` converges if and only if the improper integral `int_N^oof(x)dx` converges to a finite number.

So, we want to see if `int_1^oo1/x^2dx` converges to a finite value. When we have an improper integral such as this, replace the infinite bound with a variable, and take the limit as that variable approaches infinity. We write:

`int_1^oo1/x^2dx=lim_(brarroo)int_1^b1/x^2dx`

Then:

`=lim_(brarroo)int_1^bx^-2=lim_(brarroo) [x^-1/(-1)] _ 1^b=lim_(brarroo)[-1/x]_1^b`

Evaluating:

`=lim_(brarroo)(-1/b-(-1/1))=lim_(brarroo)(-1/b+1)`

As `brarroo`, the ratio `-1/brarr0`, so:

`=0+1=1`

So, `int_1^oo1/x^2dx=1`, that is, it converged to a finite value. Thus the series `sum_(n=1)^oo1/n^2=1+1/4+1/9+1/16+1/25+...` converges as well through the integral test.