Science:Math Exam Resources/Courses/MATH101/April 2016/Question 11 (a)

MATH101 April 2016

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Question 11 (a)
Both parts of this question concern the series $\displaystyle S=\sum _{n=1}^{\infty }(-1)^{n-1}24n^{2}e^{-n^{3}}$. (a) Show that the series $S$ converges absolutely.

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
Consider either the Integral test or the Ratio test.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Absolutly convergent series is when $\sum _{n=1}^{\infty }|f_{n}|$ converges, so we need to show that $\sum _{n=1}^{\infty }24n^{2}e^{-n^{3}}$ converges. We can apply the integral test by setting $f(x)=24x^{2}e^{-x^{3}}$, first we should check if the conditions of the integral test hold: First, $f(x)$ is certainly positive. Then we find ${\frac {df}{dx}}=(48x-72x^{4})e^{-x^{3}}=24x(2-3x^{3})e^{-x^{3}},$ which is negative for $x\geq 1$. Therefore $f(x)$ is decreasing for $x\geq 1$, and the Integral Test applies. Note that the substitution $u=x^{3}$ yields $\int f(x)\,dx=\int 24x^{2}e^{-x^{3}}\,dx=\int 8e^{-u}\,du=-8e^{-u}+C=-8e^{-x^{3}}+C.$ Therefore, ${\begin{aligned}\int _{1}^{\infty }f(x)\,dx&=\lim _{R\to \infty }\int _{1}^{R}f(x)\,dx=\lim _{R\to \infty }{\bigg (}{-}8e^{-x^{3}}{\bigg |}_{1}^{R}{\bigg )}\\&=\lim _{R\to \infty }(-8e^{-R^{3}}+8e^{-1})=8e^{-1}.\end{aligned}}$ Since the integral is convergent, $S$ converges absolutely.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Alternatively, we can use the Ratio Test. We calculate ${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {(-1)^{n}24(n+1)^{2}e^{-(n+1)^{3}}}{(-1)^{n-1}24n^{2}e^{-n^{3}}}}\right|\\&=\lim _{n\to \infty }\left({\frac {(n+1)^{2}}{n^{2}}}{\frac {e^{n^{3}}}{e^{(n+1)^{3}}}}\right)\\&=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{2}e^{-(3n^{2}+3n+1)}=1\cdot 0=0<1,\end{aligned}}$ and therefore the series converges absolutely.