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

Consider $\dfrac{n \ln n}{2^n}$ Thus, $\Sigma_{n=1}^\infty \dfrac{n \ln n}{2^n} \leq \Sigma_{n=1}^\infty \dfrac{n^2}{2^n}$ Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(n+1)^2}{2^{n+1}}}{\dfrac{n^2}{2^n}}|$ Thus, we have $l=(\dfrac{1}{2})\lim\limits_{n \to \infty}(\dfrac{n+1}{n})^2=(\dfrac{1}{2})(1)$ So, $l=\dfrac{1}{2} \lt 1$ Thus, the series $\Sigma_{n=1}^\infty \dfrac{n^2}{2^n}$ converges by the ratio test. Hence, the given series converges by the direct comparison test.