Science:Math Exam Resources/Courses/MATH101/April 2011/Question 06 (a)

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MATH101 April 2011

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Question 06 (a)
Evaluate the following indefinite integral as a power series, and find the radius of convergence. $\int {\frac {1}{1+x^{3}}}dx$

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
Remember the series expansion ${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+x^{4}+\cdots =\sum _{n=0}^{\infty }x^{n}$

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Solution

We have the series expansion ${\begin{aligned}{\frac {1}{1+x^{3}}}&={\frac {1}{1-(-x^{3})}}\\&=1+(-x^{3})+(-x^{3})^{2}+(-x^{3})^{3}+\cdots \\&=1+(-1)x^{3}+(-1)^{2}x^{6}+(-1)^{3}x^{9}+\cdots \\&=\sum _{n=0}^{\infty }(-1)^{n}x^{3n}\end{aligned}}$ and so, integrating term by term we find that ${\begin{aligned}\int {\frac {1}{1+x^{3}}}dx=\sum _{n=0}^{\infty }(-1)^{n}\int x^{3n}dx=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{3n+1}}{3n+1}}+C\end{aligned}}$ It only remains to find the radius of convergence. Computing the limit in the ratio test yields ${\begin{aligned}\lim _{n\to \infty }\left|{\frac {c_{n+1}}{c_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {(-1)^{n+1}{\frac {x^{3(n+1)+1}}{3(n+1)+1}}}{(-1)^{n}{\frac {x^{3n+1}}{3n+1}}}}\right|\\&=\lim _{n\to \infty }\left|{\frac {\frac {x^{3n+4}}{3n+4}}{\frac {x^{3n+1}}{3n+1}}}\right|\\&=\lim _{n\to \infty }{\frac {|x|^{3n+4}}{|x|^{3n+1}}}{\frac {3n+1}{3n+4}}\\&=\lim _{n\to \infty }|x|^{3}{\frac {n(3+1/n)}{n(3+4/n)}}\\&=\lim _{n\to \infty }|x|^{3}{\frac {3+1/n}{3+4/n}}\\&=|x|^{3}\end{aligned}}$ Now the ratio test tells us that the series converges whenever $\displaystyle |x|^{3}<1$ and thus our radius of convergence is 1.

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