How to determine the fourth coefficient in the power series expansion of $(1 + \ln(1-x))^{-1}$?

Question

$\dfrac{1}{1+\ln(1-x)}=\sum\limits_{n=0}^{+\infty}a_n x^n$, 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?

Answer 1

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 $\frac{11}3$.

Answer 2

Not smart at all.

In this problem, you are not asked for $a_n$ or even for $a_{123}$ but just for $a_4$.

Being lazy, what I would have done is to use $$\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}+O\left(x^6\right)$$ $$\frac{1}{1+\ln(1-x)}=\frac 1 {1-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}+O\left(x^6\right)}$$ and then long division to get $$\frac{1}{1+\ln(1-x)}=1+x+\frac{3 x^2}{2}+\frac{7 x^3}{3}+\frac{11 x^4}{3}+\frac{347 x^5}{60}+O\left(x^6\right)$$