Science:Math Exam Resources/Courses/MATH105/April 2018/Question 04 (b)/Solution 1

Let $f(x)=x^{2}e^{-3x^{3}}$. Obviously, $f(x)\geq 0$ and $f(x)$ is continuous for all $x\geq 1$. Since $f'(x)=(2x-9x^{4})e^{-3x^{3}}=x(2-9x^{3})e^{-3x^{3}}<0$ when $x\geq 1$, $f$ decreases for all $x\geq 1$. Therefore, we can apply the integral test for the series $\sum _{k=1}^{\infty }f(k)=\sum _{k=1}^{\infty }k^{2}e^{-3k^{3}}$.

To see whether the integral $\int _{1}^{\infty }f(x)dx$ converges or not, we use the substitution $u=3x^{3}$. Then, $du=9x^{2}dx$ and

$${\begin{aligned}\int _{1}^{\infty }f(x)dx&=\lim _{b\to \infty }\int _{1}^{b}x^{2}e^{-3x^{3}}dx\\&={\frac {1}{9}}\lim _{b\to \infty }\int _{3}^{3b^{3}}e^{-u}du={\frac {1}{9}}\lim _{b\to \infty }\left[-e^{-u}\right]_{3}^{3b^{3}}\\&={\frac {1}{9}}\lim _{b\to \infty }(-e^{-3b^{3}}+e^{-3})={\frac {1}{9}}e^{-3}<+\infty .\end{aligned}}$$

Since the integral converges, the series $\sum _{k=1}^{\infty }k^{2}e^{-3k^{3}}$ also converges.

Answer: $\color {blue}{\mbox{converges}}$