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

MATH105 April 2018

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Question 04 (b)
In this question, if you use the integral test for a series, please explain why the integral test can be used for the series. (b) Determine whether the series $\sum _{k=1}^{\infty }k^{2}e^{-3k^{3}}$ converges or diverges.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the integral test.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\color {blue}{\mbox{converges}}$