Find the power series of $f(x)=\frac{x}{x+1}$ in $x_0=0$

Question

In 1st Semester Calculus book I found an exercise that asks me to find the above power series of the function at the point $x_0 = 0$ using the geometric series formula and the Cauchy-Product. So far I've calculated:

$$ f(x)= \frac{x}{x+1} = \frac{\frac{x}{x}}{\frac{x+1}{x}}= \frac{1}{1+\frac{1}{x}}$$ Then with the use of the geometric series formula I've calculated:

because $$\sum_{n=0}^{\infty}{q^n}= \frac{1}{1-q} $$ with $q=(-\frac{1}{x})$ that

$$ f(x)=\sum_{n=0}^{\infty}{(-\frac{1}{x})^n}$$ and then by setting $n=-k$

$$ \sum_{k=0}^{\infty}{(-1)^{-k}x^k} $$ I've ended up with the above series, which I would say that looks like a power sum but I am not sure if it is the correct answer due to the fact that I did not use the Cauchy-Product. Therefore any insight on the problem would be appreciated.

Edit: Some corrections as stated in the comments. The way of setting n=-k is apparently wrong cause it gives an entirely different sum.

Answer 1

What you wrote is not a power series.

Note that$$\frac x{x+1}=x\times(1-x+x^2-x^2+\cdots)=x-x^2+x^3-x^4+\cdots$$if $\lvert x\rvert<1$.

Answer 2

for $| x|<1$ $$\frac{x}{x+1}=1-\frac{1}{1+x}=1-(1-x+x^2-x^3+...)$$

Answer 3

You can straight-forwardly use the geometric series :

$$\frac{1}{1+w} = \sum_{n =0}^\infty (-1)^nw^n, \; |w| < 1$$

You are expanding around zero so you have that $|x| <1 $ and simply enough, it is :

$$\frac{x}{1+x} = x\sum_{n=0}^\infty (-1)^nx^n = \sum_{n=0}^\infty (-1)^nx^{n+1}$$