Question

How do you use the integral test to determine if `Sigma n/(n^4+1)` from `[1,oo)` is convergent or divergent?

Answer 1

The series `sum_(n=N)^ooa_n` can be tested via the integral test if `a_n=f(x)` is:

• positive on `x in[N,oo)`
• decreasing on `x in [N,oo)`

We see that `f(x)=x/(x^4+1)` fits both of these on `x in [1,oo)`. The integral test then states that if the integral

`int_N^oof(x)dx`

is finite, or if the interval converges to any value, then the series `sum_(n=N)^ooa_n` converges as well.

If the integral diverges, so does the series.

So, we want to test the convergence of the integral:

`int_1^oox/(x^4+1)dx=1/2int_1^oo(2x)/((x^2)^2+1)dx`

Let `u=x^2`:

`color(white)(int_1^oox/(x^4+1)dx)=1/2int_1^oo1/(u^2+1)du`

`color(white)(int_1^oox/(x^4+1)dx)=1/2arctan(u)|_1^oo`

Note that `lim_(urarroo)arctan(u)=pi/2` and `arctan(1)=pi/4`:

`color(white)(int_1^oox/(x^4+1)dx)=1/2(pi/2-pi/4)=pi/8`

Since this integral converges, so does the given series via the integral test.