Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 12

1/4+2/7+3/12+.....+n/(n^2+3)+........
We can write the series as sum_(n=1)^oon/(n^2+3)
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^2+3)
Consider f(x)=x/(x^2+3)
Refer to the attached graph of the function. From the graph we observe that the function is positive, continuous and decreasing on the interval [1,oo)
We can also determine whether function is decreasing by finding the derivative f'(x) such that f'(x)<0 for x>=1
Now let's determine whether the corresponding improper integral int_1^oox/(x^2+3)dx converges or diverges.
int_1^oox/(x^2+3)dx=lim_(b->oo)int_1^bx/(x^2+3)dx
Let's first evaluate the indefinite integral intx/(x^2+3)dx
Apply integral substitution:u=x^2+3
=>du=2xdx
intx/(x^2+3)dx=int1/u(du)/2
Take the constant out and use the common integral:int1/xdx=ln|x|
=1/2ln|u|+C where C is a constant
Substitute back u=x^2+3
=1/2ln|x^2+3|+C
int_1^oox/(x^2+3)dx=lim_(b->oo)[1/2ln|x^2+3|]_1^b
=lim_(b->oo)1/2ln|b^2+3|-1/2ln|1^2+3|
=oo-1/2ln4
=oo
Since the integral int_1^oox/(x^2+3)dx diverges, we can conclude from the integral test that the series diverges.