Science:Math Exam Resources/Courses/MATH101 C/April 2025/Question 04/Solution 1

Apply the Ratio Test to the given series by considering the limit: $$\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{A_{n+1}}{A_n}\right|,$$ where $A_n=\frac{4^n}{\sqrt{n}}(x-6{)}^n$. Then we have

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{A_{n+1}}{A_n}\right|& =\underset{n\to \mathrm{\infty }}{\lim }|\frac{4^{n+1}}{\sqrt{n+1}}(x-6{)}^{n+1}\cdot \frac{\sqrt{n}}{4^n}\frac{1}{(x-6{)}^n}|\\ & =\underset{n\to \mathrm{\infty }}{\lim }|\frac{4^{n+1}}{4^n}\cdot \frac{\sqrt{n}}{\sqrt{n+1}}\cdot \frac{(x-6{)}^{n+1}}{(x-6{)}^n}|\\ & =\underset{n\to \mathrm{\infty }}{\lim }|4\cdot \sqrt{\frac{n}{n+1}}\cdot (x-6)|\\ & =4|x-6|\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\frac{n}{n+1}}\\ .\end{array}$$ To compute the limit, we can factor out $n$ from both the numerator and denominator, and recall $\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}=0$: $$\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\frac{n}{n+1}}=\underset{n\to \mathrm{\infty }}{\lim }\sqrt{\frac{1}{1+\frac{1}{n}}}=\sqrt{\frac{1}{1+\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}}}=1$$ Thus we have, $$\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{A_{n+1}}{A_n}\right|=4|x-6|.$$

For convergence of the power series we want

$$4|x-6|<1,\mathrm{i}\mathrm{.}\mathrm{e}\mathrm{.},|x-6|<1/4.$$ Thus, the radius of convergence is $\frac{1}{4}$. Thus, ratio of convergence is given by $1/3$.