Science:Math Exam Resources/Courses/MATH103/April 2016/Question 07 (b) (i)/Solution 1

Let $a_{n}={\frac {2^{-n}}{\sqrt {n}}}x^{n}$ and apply the ratio test:

$\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=\lim _{n\to \infty }\left|{\frac {{\frac {2^{-(n+1)}}{\sqrt {n+1}}}x^{n+1}}{{\frac {2^{-n}}{\sqrt {n}}}x^{n}}}\right|=\lim _{n\to \infty }\left|{\frac {2^{-(n+1)}}{2^{-n}}}\cdot {\frac {\sqrt {n}}{\sqrt {n+1}}}\cdot {\frac {x^{n+1}}{x^{n}}}\right|={\frac {1}{2}}|x|\lim _{n\to \infty }\left|{\frac {\sqrt {n}}{\sqrt {n+1}}}\right|={\frac {1}{2}}|x|.$

Therefore, the given series $\sum _{n=1}^{\infty }{\frac {2^{-n}}{\sqrt {n}}}x^{n}$ converges when ${\frac {1}{2}}|x|<1$ (i.e., $|x|<2$) and diverges when ${\frac {1}{2}}|x|>1$ (i.e., $|x|>2$). In other words, the radius of convergence is $\color {blue}2.$