How to find Taylor series when $x_0=0$ and radius of convergence for $\frac{x}{1+x}$ for $f:(-1,\infty)$ Through the taylor series formula:
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\dots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
I've got that $f(x)=x-x^2+x^3-x^3\dots$ however my teacher claimed that I couldn't assume that $f(x)$ exist, or something like that. So my first question is, how can I prove that $f(x)$ exists for when $x_0=0$?
Second part is concerning the radius of convergence.
I came up with the taylor series sum to be
$$\sum_{n=0}^{\infty}{(-1)^n}*x^{n+1}$$
Am I correct with that assumption, and how can I find the radius of convergence through that?
 A: If you want a series around $\;x_0=0\;$ (i.e., a Maclaurin series), then for $\;|x|<1\;$ you have
$$\frac x{1+x}=x\left(1-x+x^2-x^3+\ldots\right)=\sum_{n=0}^\infty(-1)^nx^{n+1}$$
Now if you want the series valid for $\;x\in (-1,\infty)\;$ (this is what I am guessing you meant when you wrote "for $\;f:(-1,\infty)$) , then for $\;|x|>1\iff\frac1{|x|}<1\;$ you have
$$f(x)=\frac x{1+x}=\frac1{1+\frac1x}=\sum_{n=0}^\infty\frac{(-1)^n}{x^n}$$
Thus, youy have a split definition for the series that express $\;f\;$ in different domains:
$$f(x)=\begin{cases}\sum\limits_{n=0}^\infty(-1)^nx^{n+1}\;,&\;x\in(-1,1)\\{}\\\frac12,\,&x=1\\{}\\\sum_{n=0}^\infty\frac{(-1)^n}{x^n},\,&x>1\end{cases}$$
A: First consider the finite geometric series $$S=\sum_{k=0}^{n-1}(x-x_0)^{k}=1+(x-x_0)+(x-x_0)^2...$$
Then $S-(x-x_0)S=1-(x-x_0)^n$ and $$\sum_{k=0}^{n-1}(x-x_0)^{n}=\frac{1-(x-x_0)^n}{1-(x-x_0)}$$ and we take the $lim_{n\rightarrow \infty}$
Which you can see that $(x-x_0)^n$ converges when $|x-x_0|<1$
And so if $x_0=0 \Rightarrow |x|<1$
. So for $\frac{x}{1+x}=x\cdot\frac{1}{1-(-x)}$, $x\in (-1,1)$ the expression $$\sum_{k=0}^{\infty}(-1)^k x^{k+1}$$
is valid
If you want to make the domain $(1,\infty)$ also included, you can do a simple trick,
$$\frac{x}{1+x}=\frac{1}{1+\frac{1}{x}}$$
Realize that $\left|\frac{1}{x}\right|<1$ for $|x|>1$
And so
$$\sum_{k=0}^{\infty}\frac{(-1)^k}{x^k}$$ is valid for $|x|>1$ which also includes $(1,\infty)$
So what we have done is found both the representation for $x\in(-1,1)$ and $x\in(1,\infty)$, Lets put them together
$$\frac{x}{1+x}=\begin{cases}{\sum_{k=0}^{\infty}(-1)^k x^{k+1}},x\in(-1,1)\\{}\\{\sum_{k=0}^{\infty}\frac{(-1)^k}{x^k}},x\in(1,\infty)\end{cases}$$
For $x=1$ there is no expansion since it is unincluded in $x\in(-1,1)\cup(1,\infty)$
So to answer your question, you can prove $f(x)$ exist as your expression when $x\in (-1,1)$, where it doesn't diverge.
Sorry if my explanation is bad, this is my first Answer
