How do I calculate the interval of convergence of $$ \frac{1+x}{1-x} $$
I made it into a taylor series expansion using first principles and the sum is this $$\sum_{n=-\infty}^\infty \left( \{ \begin{gather} 1\ when\ n=0\\ 2\ when\ n>0 \end{gather}\right) x^n$$ and afterwards I tried using the ratio test for $ \sum_{n=1}^\infty 2x^n$ $$\lim_{n\rightarrow\infty}\left|\frac{2x^{(n+1)}}{2x^n}\right|=|x| $$ afterwards can I just say the interval of convergence is $|x|<1$ ?? but I used wolfram alpha and it said the series diverges. So did I do it correctly? Or am I missing something?
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$\begingroup$I'm not quite sure what you mean by "first principles," but here's an approach you can take.
To start with, note that $$f(x):=\frac{1+x}{1-x}=\frac{2-1+x}{1-x}=\frac{2-(1-x)}{1-x}=\frac2{1-x}-1.$$ It is readily shown that $f(0)=1,$ and that for $n\ge 1,$ we have $$f^{(n)}(0)=n!\cdot 2,$$ so that the Taylor series for $f(x)$ about $x=0$ is given by $$\sum_{n=0}^\infty a_n x^n,$$ where $$a_n=\begin{cases}1 & n=0\\2 & n\ge 1.\end{cases}$$ Then we have $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac22\right|=1,$$ and so the radius of convergence is $\frac11=1,$ as you claimed. Note that Wolfram|Alpha confirms convergence when $|x|<1.$
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