Science:Math Exam Resources/Courses/MATH101/April 2015/Question 06 (b)/Solution 1

Since the series $\sum _{n=0}^{\mathrm{\infty }}(1-b_n)$ converges, we have

$\underset{n\to \mathrm{\infty }}{\lim }(1-b_n)=0⟺\underset{n\to \mathrm{\infty }}{\lim }{b}_n=1$.

Applying the ratio test, the series $\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^n$ converges when

$\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{b_{n+1}{x}^{n+1}}{b_n{x}^n}\right|=\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{b_{n+1}}{b_n}\right||x|=\left|\frac{\underset{n\to \mathrm{\infty }}{\lim }{b}_{n+1}}{\underset{n\to \mathrm{\infty }}{\lim }{b}_n}\right||x|=|x|<1,$

and diverges when $|x|>1$. The last equality follows from $\underset{n\to \mathrm{\infty }}{\lim }{b}_n=1⟹\underset{n\to \mathrm{\infty }}{\lim }{b}_{n+1}=1.$

Therefore, the radius of convergence of the power series $\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^n$ is equal to $1$.