## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=(\dfrac{n}{3n+1})^n$ When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase. Thus, we have $a_n \leq (\dfrac{n}{3n})^n=(\dfrac{1}{3})^n$ Here, $\Sigma_{n=1}^\infty (\dfrac{1}{3})^n$ is a convergent series due to the comparison test.