Science:Math Exam Resources/Courses/MATH101/April 2017/Question 11 (a)/Solution 1

We can easily see that when $x=2$,

$\sum _{n=0}^{\infty }A_{n}(x-1)^{n}=\sum _{n=0}^{\infty }A_{n}$.

Since the radius of convergence is given as $R={\frac {3}{2}}$ for the power series $\sum _{n=0}^{\infty }A_{n}(x-1)^{n}$, the series converges when $|x-1|<{\frac {3}{2}}$.

Therefore, $|2-1|=1<{\frac {3}{2}}$ implies the convergence of the series $\sum _{n=0}^{\infty }A_{n}$.

Answer: $\color {blue}\sum _{n=0}^{\infty }A_{n}\ {\text{converges}}.$