Calculating lagrange error term I have the function $ \ln \sqrt{\frac{1+x}{1-x}} $
I found its maclaurin series by using the known series for $\ln (1+x)$ and then approximated $\ln 2$ using the first three terms of this series.
How do I find the lagrange error term now? Do I have to go through the derivatives of $ \ln \sqrt{\frac{1+x}{1-x}} $ or is there an easier way?
 A: Let $f(x)=\log\left(\sqrt{\frac{1+x}{1-x}}\right)$.  We can develop the series expansion of $f(x)$ by using the series expansions for $\log(1+x)$ and $\log(1-x)$.
These are respectively
$$\begin{align}
\log(1+x)&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}\\\\\log(1-x)&=-\sum_{n=1}^\infty \frac{x^n}{n}
\end{align}$$
for $|x|<1$, whereby we see that 
$$\begin{align}
f(x)&=\log\left(\sqrt{\frac{1+x}{1-x}}\right)\\\\
&=\frac12\left(\log(1+x)-\log(1-x)\right)\\\\&=\frac12\sum_{n=1}^\infty \frac{((-1)^{n-1}+1)x^n}{n}\\\\
&=\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}\tag 1
\end{align}$$
Using the series in $(1)$, we see that $\log(2)=\log\left(\sqrt{\frac{1+3/5}{1-3/5}}\right)=\sum_{n=0}^\infty \frac{(3/5)^{2n+1}}{2n+1}$.
The Lagrange form of the remainder of the series is given by
$$R_n=\frac{f^{(n+1)}(\xi)x^{n+1}}{(n+1)!}$$
The $n$th derivative of $f(x)$ is trivial using $f(x)=\frac12\left(\log(1+x)-\log(1-x)\right)$.  Then, we have
$$f^{(n+1)}(\xi)=\frac{n!}{2}\left(\frac{(-1)^{n}}{(1+\xi)^{n+1}}+\frac{1}{(1-\xi)^{n+1}}\right)$$
