<s>[INST] How to determine the fourth coefficient in the power series expansion of $(1 + \ln(1-x))^{-1}$? 
$\dfrac{1}{1+\ln(1-x)}=\sum\limits_{n=0}^{+\infty}a_n x^n$, then $a_4=$?

My approach:

\begin{align*}\dfrac{1}{1+\ln(1-x)}&=\sum\limits_{n=0}^{+\infty}(-1)^n(\ln(1-x))^n\\
&=\sum\limits_{n=0}^{+\infty}(-1)^n\left(\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\end{align*}
Then $$a_4=(-1)^1\cdot\dfrac{1}{4}+(-1)^2\cdot\left(\dfrac{1}{2}\cdot \dfrac{1}{2}+\binom{2}{1}\cdot 1\cdot\dfrac{1}{3}\right)+ (-1)^3 \cdot\binom{3}{1}\cdot 1\cdot1\cdot\dfrac{1}{2}+ (-1)^4\cdot 1\cdot1\cdot 1\cdot 1=\dfrac{1}{6}$$

But the answer is $\dfrac{11}{3}$, ignoring $(-1)^n$. What’s the problem?
 [/INST] You left out a negative sign in the logarithmic series:
$$\begin{align*}
  \dfrac{1}{1+\ln(1-x)}
  &=\sum\limits_{n=0}^{+\infty}(-1)^n(\ln(1-x))^n\\
  &=\sum\limits_{n=0}^{+\infty}(-1)^n
    \left(-\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\\
  &=\sum\limits_{n=0}^{+\infty}
    \left(\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\end{align*}$$
and expanding this carefully gives the coefficient of $x^4$ as $\frac{11}3$.
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