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

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

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

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

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

Answer: $\sum _{n=0}^{\mathrm{\infty }}{A}_n\text{converges}.$