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

We have the series expansion

$\begin{array}{rl}\frac{1}{1+x^3}& =\frac{1}{1-(-x^3)}\\ & =1+(-x^3)+(-x^3{)}^2+(-x^3{)}^3+\cdots \\ & =1+(-1)x^3+(-1{)}^2{x}^6+(-1{)}^3{x}^9+\cdots \\ & =\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{x}^{3n}\end{array}$

and so, integrating term by term we find that

$\begin{array}{r}\int \frac{1}{1+x^3}dx=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\int x^{3n}dx=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{x^{3n+1}}{3n+1}+C\end{array}$

It only remains to find the radius of convergence. Computing the limit in the ratio test yields

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{c_{n+1}}{c_n}\right|& =\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{(-1{)}^{n+1}\frac{x^{3(n+1)+1}}{3(n+1)+1}}{(-1{)}^n\frac{x^{3n+1}}{3n+1}}\right|\\ & =\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{\frac{x^{3n+4}}{3n+4}}{\frac{x^{3n+1}}{3n+1}}\right|\\ & =\underset{n\to \mathrm{\infty }}{\lim }\frac{|x{|}^{3n+4}}{|x{|}^{3n+1}}\frac{3n+1}{3n+4}\\ & =\underset{n\to \mathrm{\infty }}{\lim }|x{|}^3\frac{n(3+1/n)}{n(3+4/n)}\\ & =\underset{n\to \mathrm{\infty }}{\lim }|x{|}^3\frac{3+1/n}{3+4/n}\\ & =|x{|}^3\end{array}$

Now the ratio test tells us that the series converges whenever $|x{|}^3<1$ and thus our radius of convergence is 1.