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

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

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

For convergence of the power series we want

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