Calculus of a Single Variable, Chapter 9, 9.3, Section 9.3, Problem 2

sum_(n=1)^oo2/(3n+5)
The Integral test is applicable if f is positive, continuous and decreasing function on the infinite interval [k,oo) where k>=1 and a_n=f(x) . Then the series converges or diverges if and only if the improper integral int_k^oof(x)dx converges or diverges.
For the given series a_n=2/(3n+5)
Consider f(x)=2/(3x+5)
Refer to the attached graph of the function. It is positive and continuous on the interval [1,oo)
We can determine whether f(x) is decreasing by finding the derivativef'(x) , f'(x)<0 for x>=1
f'(x)=2(-1)(3x+5)^(-2)(3)
f'(x)=-6/(3x+5)^2
So,f'(x)<0
We can apply integral test as the function satisfies the conditions for the integral test.
Now let's determine whether the corresponding improper integral int_1^oo2/(3x+5)dx converges or diverges as:
int_1^oo2/(3x+5)dx=lim_(b->oo)int_1^b2/(3x+5)dx
=lim_(b->oo)[2/3ln|3x+5|]_1^b
=lim_(b->oo)2/3[ln|3b+5|-ln|3+5|]
=2/3[oo-ln8]
=oo which implies that the integral diverges.
Therefore the series sum_(n=1)^oo2/(3n+5) must also diverge.