<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] 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.
</s>