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

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

To see whether the integral $\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)dx$ converges or not, we use the substitution $u=3x^3$. Then, $du=9x^{2}dx$ and $$\begin{array}{rl}\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)dx& =\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}{x}^2{e}^{-3x^3}dx\\ & =\frac{1}{9}\underset{b\to \mathrm{\infty }}{\lim }\underset{3}{\overset{3b^3}{\int }}{e}^{-u}du=\frac{1}{9}\underset{b\to \mathrm{\infty }}{\lim }{[-e^{-u}]}_3^{3b^3}\\ & =\frac{1}{9}\underset{b\to \mathrm{\infty }}{\lim }(-e^{-3b^3}+e^{-3})=\frac{1}{9}{e}^{-3}<+\mathrm{\infty }.\end{array}$$

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

Answer: $\text{converges}$