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.
 $\int {\frac {1}{1+x^{3}}}dx$

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Hint

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

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Solution

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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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