Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$

Question

Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$ then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following equation for $x$: $$1-x=(1-u)^2$$ $$x=u(2-u)$$ and replacing $u$ by and $x$, and $=$ by $\to$ , which leads to $$\frac{1}{(1-x)^2}=1+x(2-x)+x^2(2-x)^2+x^3(2-x)^3+... \tag{2}$$

---

Succumbing to the same 'substitution'approach with $\frac{x}{1+x^2}$, I put $$\frac{1}{1-x}=\frac{u}{1+u^2}$$ $$x=1-\frac{1}{u}-u$$

so with $x \to 1-\frac{1}{x}-x$

I should have: $$\frac{x}{1+x^2}=1+\left(1-\frac{1}{x}-x\right)+\left(1-\frac{1}{x}-x\right)^2+\left(1-\frac{1}{x}-x\right)^3+...\tag{3}$$ but this does not seem correct (at least according to my Desmos graph for $|x|<1$).

Can someone please explain what my conceptual errors are?

Answer 1

Your series $(1)$ is valid for $x\in(-1,1)$.

So series $(2)$ is valid for $x(2-x)\in(-1,1)$, i.e. for $x\in(1-\sqrt 2,1)\cup(1,1+\sqrt 2)$.

But series $(3)$ is only valid if $1-\frac{1}{x}-x\in(-1,1)$. And it turns out that this is not true for any $x\in\Bbb R$.

Answer 2

Hint:

Replace $x$ with $-y^2$ to find

$$\dfrac1{1-(-y^2)}=\sum_{r=0}^\infty(-y^2)^r$$

Answer 3

$$\frac{x}{1+x^2}=\frac{1}{1-(\color{red}{1-x-\frac 1x})}$$

$$=\frac{1}{1-u}$$

When $ x \to 0 $, $ u =\color{red}{1-x-\frac 1x} $ goes to $ \infty $, so you are not allowed to repalce $ X $ by $ u $ in the expansion $$\frac{1}{1-X}=1+X+X^2+..$$

In the first part, $ u=x(2-x)$ goes to zero.

Answer 4

Because

$$u^2+1=ux(1-x)$$

$$u^2-ux(1-x)+1=0$$

has no real solutions.