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

Question

6671 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-05T05:31:40

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

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.