<s>[INST] Maclaurin series of $\ln \left( 1+\frac{\ln(1+x)}{1+x} \right)$ So the first thing I done was
$$\begin{align}\ln(1+x)&=x-\frac{1}{2}x^2+\frac{1}{3}x^3+o(x^3)\\&=(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)\end{align}$$
I've never seen this done but I'm pretty sure I can do this.
Now I want to divide:
$$(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)$$By $(1+x)$ but I don't know what
$\frac{o(x^3)}{1+x}$ will be.
Also is there another way to do this?
 [/INST] Let $u=1+x$ and $w=u+\log(u$).  Rewrite
$$
f(x)=\log(1+\frac{\log(1+x)}{1+x})=\log(w) -\log(u)
$$
We seek the coefficients in the Taylor series expansion
$$
f(x) = \sum_{n=0}^{\infty}f^{\{n\}}(0)\frac{x^n}{n!}
$$
where the coefficient $f^{\{n\}}(0)$ is the nth derivative of the function evaluated at $x=0$. We find
$$
f^{'}(x)=\frac{w^{'}}{w}-\frac{1}{u}
$$
$$
f^{"}(x)=\frac{w^{"}}{w}-\frac{w^{'}}{w^2} +\frac{1}{u^2}
$$
$$
f^{\{3\}}(x)=\frac{w^{\{3\}}}{w}-\frac{2w^{"}}{w^2} +\frac{2w^{'}}{w^3}-\frac{2}{u^3}
$$
$$
f^{\{4\}}(x)=\frac{w^{\{4\}}}{w}- \frac{3w^{\{3\}}}{w^2}+\frac{6w^{"}}{w^3} -\frac{6w^{'}}{w^4}+\frac{6}{u^4}
$$
$$
f^{\{5\}}(x)=\frac{w^{\{5\}}}{w}- \frac{4w^{\{4\}}}{w^2}+\frac{12w^{\{3\}}}{w^3} -\frac{24w^{"}}{w^4}+\frac{24w^{'}}{w^5}-\frac{24}{u^5}
$$
We now observe
$$
w(0)=1\\
w^{'}(0)=2\\
w^{"}(0)=-1\\
w^{\{3\}}(0)=2\\
w^{\{4\}}(0)=-6=-3!\\
w^{\{5\}}(0)=24=4!
$$
Plugging these values into our derivative equations we find by induction,
$$
f(0)=0\\
f^{'}(0)=1\\
f^{"}(0)=-2\\
f^{\{3\}}(0)=6=3!\\
f^{\{4\}}(0)=-24=-4!\\
f^{\{5\}}(0)=5!\\
.\\
.\\
.\\
f^{\{n\}}(0)= (-1)^{n+1}n!\\
$$
Thus, interestingly enough,
$$
\log(1+\frac{\log(1+x)}{1+x})=\sum_{n=1}^{\infty}(-1)^{n+1}x^n
$$
</s>