Answer
convergent
Work Step by Step
Let us consider $f(x)=\dfrac{x^2}{e^{x/3}}$
Here, the function $f(x)$ is positive, continuous for $x \geq 1$ and $f(x)$ is decreasing for $x \gt 7$
Then $\int_7^\infty \dfrac{x^2}{e^{x/3}}dx= \lim\limits_{k \to \infty} \int_7^k \dfrac{x^2}{e^{x/3}}dx$
or, $ \lim\limits_{k \to \infty} [\dfrac{-54}{e^{k/3}}+\dfrac{327}{e^{7/3}}]=\dfrac{327}{e^{7/3}}$
Thus, the sequence $\Sigma_{n=1}^\infty \dfrac{n^2}{e^{n/3}}$ is convergent.
