I'm trying to find a Taylor Series representation of $f(x)=\ln(\frac{1+2x}{1-2x})$ centered at $0$. So I am using the Maclaurin Series representation of $f(x)=\ln(1+x)$ which is $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$. Then I switch up the original function and plug in my new values of $x$ in the the Maclaurin Series of $\ln(1+x)$. $$f(x)=\ln\left(\frac{1+2x}{1-2x}\right)=f(x)=\ln\left(\frac{1+2x}{1+(-2x)}\right)=\ln(1+2x)-\ln(1+(-2x))$$ $$f(x)=\ln\left(\frac{1+2x}{1-2x}\right)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2x)^n}{n}-\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(-2x)^n}{n}$$ $$= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}(2x)^n-(-1)^{n-1} (-2x)^n}{n} $$ $$= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\left( (2x)^n- (-2x)^n\right)}{n} $$ $$= \sum_{n=0}^{\infty}\frac{(-1)^{n}\left( (2x)^{n+1}- (-2x)^{n+1}\right)}{n+1} $$
And that was my final answer, but it's wrong. How do I go about finding this? Thanks for all the help in advance.
In what sense is it wrong?
$\displaystyle \frac{(-1)^{n}\left( (2x)^{n+1}- (-2x)^{n+1}\right)}{n+1}$ is zero when $n+1$ is even, i.e. when $n$ is odd.
Meanwhile, when $n$ is even $(-1)^{n}=1$ and $(2x)^{n+1}=- (-2x)^{n+1}$ so $\displaystyle \frac{(-1)^{n}\left( (2x)^{n+1}- (-2x)^{n+1}\right)}{n+1} = \frac{2^{n+2}x^{n+1}}{n+1}$ which means you could rewrite your answer as $$ \sum_{n=0}^{\infty}\frac{2^{2n+2}x^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{2}{{2n+1}}(2x)^{2n+1}$$ or numerous other possibilities. The problem may be that computerised marking of such expressions is not an exact science.