## Thomas' Calculus 13th Edition

Let $u_n=\dfrac{1}{n3^{n}}$ When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing. $\implies a_n \leq \dfrac{1}{3^{n}}$ we can see that the series $\Sigma_{n=1}^\infty \dfrac{1}{3^n}$ shows a convergent geometric series and has common ratio $r=\dfrac{1}{3} \gt 1$ . Hence, the series converges by the comparison test.