Answer
(a) If the power series $ \Sigma_{n=0}^\infty c_n(x-a)^n$(centered at $a$) has radius of convergence $R$ , then the function $f$ that represents the series can be written as: $f(x)= c_0+c_1(x-a)+c_2(x-1)^2+...+c_n(x-a)^n$. The function is differnaiable on the interval $(a-R,a+R)$, and $f'(x)= c_1(x-a)+2c_2(x-a)+...+nc_n(x-a)^{n-1}$.$f'(x)$ has radius of convergence $R$.
(b) If the power series $ \Sigma_{n=0}^\infty c_n(x-a)^n$(centered at $a$) has radius of convergence $R$ , then the function $f$ that represents the series can be written as: $f(x)= c_0+c_1(x-a)+c_2(x-1)^2+...+c_n(x-a)^n$. The integral of this function is $F(x)=\int f(x) dx= C+c_0(x-a)+c_1(x-a)^2/2+c_2(x-a)^3/3+...+c_n(x-a)^{n+1}/(n+1)$.
$F(x)$ has radius of convergence $R$.
Work Step by Step
(a) If the power series $ \Sigma_{n=0}^\infty c_n(x-a)^n$(centered at $a$) has radius of convergence $R$ , then the function $f$ that represents the series can be written as: $f(x)= c_0+c_1(x-a)+c_2(x-1)^2+...+c_n(x-a)^n$. The function is differnaiable on the interval $(a-R,a+R)$, and $f'(x)= c_1(x-a)+2c_2(x-a)+...+nc_n(x-a)^{n-1}$.$f'(x)$ has radius of convergence $R$.
(b) If the power series $ \Sigma_{n=0}^\infty c_n(x-a)^n$(centered at $a$) has radius of convergence $R$ , then the function $f$ that represents the series can be written as: $f(x)= c_0+c_1(x-a)+c_2(x-1)^2+...+c_n(x-a)^n$. The integral of this function is $F(x)=\int f(x) dx= C+c_0(x-a)+c_1(x-a)^2/2+c_2(x-a)^3/3+...+c_n(x-a)^{n+1}/(n+1)$.
$F(x)$ has radius of convergence $R$.
