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

Consider $f(x)=\dfrac{x^2}{e^{x/3}}$ This is positive, continuous for $x \geq 1$ and $f(x)$ is decreasing for $x \gt 7$ Now, take the integral test to find the convergence and divergence for the sequence. We have $\int_7^\infty \dfrac{x^2}{e^{x/3}}dx= \lim\limits_{a \to \infty} \int_7^a \dfrac{x^2}{e^{x/3}}dx$ or, $\lim\limits_{a \to \infty} [\dfrac{-54}{e^{b/3}}+\dfrac{327}{e^{7/3}}]=\dfrac{327}{e^{7/3}}$ Hence, the sequence $\Sigma_{n=1}^\infty \dfrac{n^2}{e^{n/3}}$ is convergent