Question

11. (5 pts) Find all values of x for which the power series converges: 1- 4(x – 3) + (x – 3) * - (x – 3)' +....

Answer 1

$f(x)=1-\frac{x-3}{2}+\frac{(x-3)^2}{3}-\frac{(x-3)^3}{4}$

$f(x)=\frac{(x-3)^0}{1}-\frac{(x-3)^1}{2}+\frac{(x-3)^2}{3}-\frac{(x-3)^3}{4}$

$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(x-3)^{n-1}}{n}$

$=>a_n=\frac{(-1)^{n-1}(x-3)^{n-1}}{n}$

$=>a_{n+1}=\frac{(-1)^{n}(x-3)^{n}}{n+1}$

Power series converges when

ant lim n-> an

$\lim_{n->\infty}|\frac{ \frac{(-1)^{n}(x-3)^{n}}{n+1} }{ \frac{(-1)^{n-1}(x-3)^{n-1}}{n} }|<1$

$\lim_{n->\infty}|\frac{ -n(x-3) }{ n+1 }|<1$

$\lim_{n->\infty}|\frac{ -(x-3) }{ 1+\frac{1}{n} }|<1$

$|\frac{ -(x-3) }{ 1+0 }|<1$