Find the Maclaurin series representation of the function: $\ln(\frac{1+x}{1-x})$ Find the Maclaurin series representation of the function: $\ln(\frac{1+x}{1-x})$ I know the answer, but i want to know why my approch did not work.

I started with the series
$$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n $$
then i put in $x=\frac{2x}{1-x}$ and get:
$$\ln(1+x)=\ln(1+\frac{2x}{1-x})=\ln(\frac{1+x}{1-x}))=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}(\frac{2x}{x-1})^n$$
It seems like this is not correct based on wolfam alpha. Where did i make a mistake?
This is not formally incorrect (although you would have to define in what sense the series converges), it is simply not a Maclaurin series. A Maclaurin series is expressed in powers of $x$ with constant coefficients, here you express it in inverse powers of $1-x$ which is not the form of the series. You could try to expand each function $\left(\frac{2x}{x-1}\right)^n$ but that is unnecessarily complicated and dangerous.
Much simpler, write $\ln \left( \frac{1+x}{1-x} \right)=\ln (1+x)-\ln (1-x)$, develop each function separately and check which terms cancel out to find the answer. I will let you conclude!
This is not formally incorrect (although you would have to define in what sense the series converges), it is simply not a Maclaurin series. A Maclaurin series is expressed in powers of $x$ with constant coefficients, here you express it in inverse powers of $1-x$ which is not the form of the series. You could try to expand each function $\left(\frac{2x}{x-1}\right)^n$ but that is unnecessarily complicated and dangerous.
Much simpler, write $\ln \left( \frac{1+x}{1-x} \right)=\ln (1+x)-\ln (1-x)$, develop each function separately and check which terms cancel out to find the answer. I will let you conclude!