# Taylor Series Maclaurin Series Interval Expansion

I am currently woking on some clack online homework problem. I really have no idea how to approach this problem. If someone could help me solve this question I would greatly appreciate it!

From Rogawski ET 2e section 10.7, exercise 31.

Find the Taylor series for $f(x) = \dfrac{1}{1 - 3x}$ centered at $c=1$.

$$\frac{1}{1-3x} = \sum_{n=0}^{\infty} [\textrm{_________}]$$

$$\begin{array} \\ \frac{1}{1-3x} &= \frac{1}{1-3(x-1)-3} \\ &= \frac{1}{-2-3(x-1)} \\ &= -\frac{1}{2} \left( \frac{1}{1 + \frac{3}{2}(x-1)} \right) \\ &= -\frac{1}{2} \left( \frac{1}{1 - (-\frac{3}{2}(x-1))} \right) \\ &= -\frac{1}{2} \sum_{n=0}^\infty \left(-\frac{3}{2}(x-1)\right)^n \textrm{ for } \left|-\frac{3}{2}(x-1)\right|<1\\ &= -\frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n 3^n}{2^n} (x-1)^n \textrm{ for } \left|x-1\right|<\frac{2}{3} \\ &= \sum_{n=0}^\infty \frac{(-1)^{n+1} 3^n}{2^{n+1}} (x-1)^n \textrm{ for } x \in \left(\frac{1}{3},\frac{5}{3}\right) \end{array}$$

Try writing out the expansion on a piece of paper. It should look like $$\frac{1}{1-3x}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(x-1)^n}{2^{n+1}}$$ Now, consider how the $k^{th}$ term and the $k+1^{th}$ term compare.

$$T_k=\frac{(-1)^{k+1}(x-1)^k}{2^{k+1}}$$ $$T_{k+1}=\frac{(-1)^{k+1+1}(x-1)^{k+1}}{2^{k+1+1}}=\frac{(-1)^{k+1}(x-1)^{k}}{2^{k+1}}\frac{(-1)(x-1)}{2}=\frac{(1-x)}{2}T_k=r(x)T_k$$ Obviously $|r(x)|<1$ must hold in order for the sum to converge, requiring $x$ to fall in the interval $(-1,3)$