<s>[INST] Maclaurin series for ln I need to find the Maclaurin series for $\ln\sqrt{\frac{1+x}{1-x}} $. I found it is $\frac{x^{2n+1}}{{2n+1}}$ ,is it correct? I got the 1/2 outside and solved the maclaurin for the normal log. If it is okay how can I find a relationship between this and the maclaurin series artg.I know it is $(-1)^n$ of my series but how can I write it?
 [/INST] The Taylor series for $\log\sqrt{\frac{1+x}{1-x}}$ at $0$ is $\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}$, because$$\begin{align}\log\sqrt{\frac{1+x}{1-x}}&=\frac12\log\left(\frac{1+x}{1-x}\right)\\&=\frac12\left(\log(1+x)-\log(1-x)\right)\\&=\frac12\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots-\left(-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4-\cdots\right)\right)\\&=x+\frac{x^3}3+\frac{x^5}5+\cdots\end{align}$$And the Taylor series for $\arctan(x)$ at $0$ is quite similar: $\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$. That's because$$\arctan'(x)=\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$$
</s>