Science:Math Exam Resources/Courses/MATH101/April 2011/Question 06 (a)/Solution 1

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.