Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 21

sum_(n=1)^oon/(n^4+1)
The integral test is applicable if f is positive, continuous and decreasing function on infinite interval [k,oo) where k>=1 and a_n=f(x) . Then the series sum_(n=1)^ooa_n converges or diverges if and only if the improper integral int_1^oof(x)dx converges or diverges.
For the given series a_n=n/(n^4+1)
Consider f(x)=x/(x^4+1)
Refer to the attached graph of the function. From the graph we can observe that the function is positive , continuous and decreasing on the interval [1,oo)
We can determine whether function is decreasing, also ,by finding the derivative f'(x) such that f'(x)<0 for x>=1 .
We can apply integral test , since the function satisfies the conditions for the integral test.
Now let's determine whether the corresponding improper integral int_1^oox/(x^4+1)dx converges or diverges.
int_1^oox/(x^4+1)dx=lim_(b->oo)int_1^bx/(x^4+1)dx
Let's first evaluate the indefinite integral intx/(x^4+1)dx
Apply integral substitution:u=x^2
=>du=2xdx
intx/(x^4+1)dx=int1/(u^2+1)(du)/2
Take the constant out and use common integral:int1/(x^2+1)dx=arctan(x)+C
=1/2arctan(u)+C
Substitute back u=x^2
=1/2arctan(x^2)+C
int_1^oox/(x^4+1)dx=lim_(b->oo)[1/2arctan(x^2)]_1^b
=lim_(b->oo)[1/2arctan(b^2)]-[1/2arctan(1^2)]
=lim_(b->oo)[1/2arctan(b^2)]-[1/2arctan(1)]
=lim_(b->oo)[1/2arctan(b^2)]-[1/2(pi/4)]
=lim_(b->oo)[1/2arctan(b^2)]-pi/8
=1/2lim_(b->oo)arctan(b^2)-pi/8
Now lim_(b->oo)(b^2)=oo
=1/2(pi/2)-pi/8 [by applying the common limit:lim_(u->oo)arctan(u)=pi/2 ]
=pi/4-pi/8
=pi/8
Since the integral int_1^oox/(x^4+1)dx converges, we conclude from the integral test that the series converges.