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?
 A: Following @jjagmaths approach we show
\begin{align*}
\color{blue}{\ln\left(1+\frac{\ln(1+x)}{1+x}\right)}
&\color{blue}{=\sum_{n=1}^\infty(-1)^{n-1}\sum_{m=1}^{n}\frac{1}{m}
\sum_{{k_1+\cdots +k_m=n}\atop{k_1,\ldots,k_m\geq 1}}H_{k_1}\cdots H_{k_m}x^n}\tag{1}\\
&=x-2x^2+\frac{11}{3}x^3\color{blue}{-\frac{163}{24}}x^4+\cdots
\end{align*}

Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain for $n\geq 1$:
\begin{align*}
[x^n]&\ln\left(1+\frac{\ln(1+x)}{1+x}\right)\\
&=\sum_{m=1}^{n}\frac{(-1)^{m-1}}{m}[x^n]\frac{\ln^m(1+x)}{(1+x)^m}\tag{2}
\end{align*}

In (2) we note that since $\ln(1+x)$ starts with $x$, the $m$-th power of $\frac{\ln(1+x)}{1+x}$ starts with $x^m$ and we can restrict the sum with the upper limit $n$. We have
\begin{align*}
A(x)&=\frac{\ln(1+x)}{1+x}\\
&=\left(\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}x^k\right)\left(\sum_{l=0}^\infty(-1)^lx^l\right)\\
&=\sum_{q=1}^\infty\left(\sum_{{k+l=q}\atop{k\geq 1, l\geq 0}}\frac{(-1)^{k-1+l}}{k}\right)x^q\\
&=\sum_{q=1}^\infty(-1)^{q-1}\left(\sum_{k=1}^q\frac{1}{k}\right)x^q\\
&=\sum_{q=1}^\infty(-1)^{q-1}H_qx^q\\
\end{align*}
It follows for $m\geq 1$:
\begin{align*}
A^m(x)&=\sum_{q=m}^{\infty}\left(\sum_{{k_1+\cdots+k_m=q}\atop{k_1,\ldots,k_m\geq 1}}
(-1)^{k_1-1}H_{k_1}\cdots(-1)^{k_m-1}H_{k_m}\right)x^q\\
&=\sum_{q=m}^{\infty}(-1)^{q-m}\sum_{{k_1+\cdots+k_m=q}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}x^q\\
\end{align*}
and for $n\geq m$:
\begin{align*}
[x^n]A^m(x)&=(-1)^{n-m}\sum_{{k_1+\cdots+k_m=n}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}\tag{3}
\end{align*}

Putting (3) in (2) we obtain
\begin{align*}
\color{blue}{[x^n]\ln\left(1+\frac{\ln(1+x)}{1+x}\right)
=\sum_{m=1}^{n}\frac{(-1)^{n-1}}{m}\sum_{{k_1+\cdots+k_m=n}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}}\tag{4}
\end{align*}
and the claim (1) follows.

Calculating for instance the coefficient of $x^4$ we obtain from (4)
\begin{align*}
\color{blue}{[x^4]}&\color{blue}{\ln\left(1+\frac{\ln(1+x)}{1+x}\right)}\\
&=\sum_{m=1}^{4}\frac{(-1)^{n-1}}{m}\sum_{{k_1+\cdots+k_m=4}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}\\
&=-\left(H_4+\frac{1}{2}\left(2H_1H_3+H_2^2\right)+\frac{1}{3}\left(3H_1^2H_2\right)+\frac{1}{4}H_1^4\right)\\
&=-\left(\frac{25}{12}+\frac{1}{2}\left(2\cdot 1\cdot \frac{11}{6}+\left(\frac{3}{2}\right)^2\right)+\frac{1}{3}\left(3\cdot 1\cdot\frac{3}{2}\right)+\frac{1}{4}\cdot 1\right)\\
&\,\,\color{blue}{=-\frac{163}{24}}
\end{align*}
in accordance with the claim (1) and Wolfram Alpha.
A: Right now I don't have time to post the complete solution, but the series is:
$$\sum_{n=1}^\infty (-1)^{n+1}\left(\sum_{k_1+k_2+\cdots+k_m=n}\frac{1}{m}H_{k_1}H_{k_2}\cdots H_{k_m}\right)x^n$$
where $H_k$ is the $k$-th harmonic number and the inner sum is taken over all the possible ways of writing $n$ as a sum of positive integers.
A: 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,
$$
\log(1+\frac{\log(1+x)}{1+x})=\sum_{n=1}^{\infty}(-1)^{n+1}x^n
$$
