How do you use the integral test to determine if #Sigma n/(n^4+1)# from #[1,oo)# is convergent or divergent?
1 Answer
The series
- positive on
#x in[N,oo)# - decreasing on
#x in [N,oo)#
We see that
#int_N^oof(x)dx#
is finite, or if the interval converges to any value, then the series
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
#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
#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.