How to find the Taylor series of order $3$ at $x=0$ of the function $$f(x)=\frac{\ln(1+x)}{(e^x-1)}.$$
The general strategy is to lift the Taylor series of $1/(e^x-1)$ up and multiply it with $\ln (1+x)$ but I am not able to do that partly because I am trying to use the form $$\frac{1}{1-u(x)}=1+(u(x))^2+(u(x))^3+....$$ but in order for this to work I need $u(x)\to0$ as $x\to 0.$ This is not the case here as $e^x\to 1$ as $x\to 0.$ So any hints/insights will be much appreciated.
On
$$ \frac{e^x-1}{x} = 1 + \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + O(x^4) $$ $$ ( 1 + ax ) \frac{e^x-1}{x} = 1 + (a + \frac{1}{2}) x + O(x^2) $$ $$ ( 1 - \frac{x}{2} + b x^2 ) \frac{e^x-1}{x} = 1 + (b - \frac{1}{12}) x^2 + O(x^3) $$ $$ ( 1 - \frac{x}{2} + \frac{ x^2}{12} ) \frac{e^x-1}{x} = 1 + O(x^4) $$ This was a surprise, the cube term disappeared $$ \frac{x}{e^x-1} = 1 - \frac{x}{2} + \frac{ x^2}{12} + O(x^4) $$ Now multiply by $\frac{\log(1+x)}{x}$
On
From $\frac{\log(1+x)}{x}=\sum_{n\geq 0}\frac{(-1)^{n}}{n+1}x^n$ and $\frac{x}{e^x-1}=\sum_{n\geq 0}\frac{B_n}{n!}x^n $ we have
$$ \frac{\log(1+x)}{e^x-1} = \sum_{m\geq 0}(-1)^m x^m\sum_{n=0}^{m}\frac{B_n(-1)^n}{n!(m-n+1)}=1-x+\frac{2}{3}x^2-\frac{11}{24}x^3+O(x^4). $$
Simpler: consider the expansion at order $4$ of the numerator and denominator (there will be a simplification by $x$): $$\frac{\ln(1+x)}{\mathrm e^x-1}=\frac{x-\cfrac{x^2}2+\cfrac{x^3}3-\cfrac{x^4}4+o(x^4)}{x+\cfrac{x^2}2+\cfrac{x^3}6+\cfrac{x^4}{24}+o(x^4)}= \frac{1-\cfrac{x}2+\cfrac{x^2}3-\cfrac{x^3}4+o(x^3)}{1+\cfrac{x}2+\cfrac{x^2}6+\cfrac{x^3}{24}+o(x^3)}$$ and perform the division by increasing powers, up to order $3$, of $\;1-\cfrac{x}2+\cfrac{x^2}3-\cfrac{x^3}4 \;$ by $\;1+\cfrac{x}2+\cfrac{x^2}6+\cfrac{x^3}{24} $.