maclaurin series for $\frac{x}{\ln(1+x)}$.

Question

we know that the Maclaurin series for $\ln(1+x)$ is $-\sum\limits_k\frac{(-1)^kx^k}{k}$. I am having a hard time getting the Maclaurin series for $\frac{x}{\ln(1+x)}$ since when $x=0$, I get $\frac{0}{0}$.

Any help is very much appreciated.

Answer 1

Be $\begin{bmatrix}k\\j\end{bmatrix}$ the unsigned Stirling numbers of the first kind.

Then with $|x|<1$ and $\displaystyle x=e^{\ln(1+x)}-1$ one gets

$\displaystyle \frac{x}{\ln(1+x)}=\sum\limits_{k=0}^\infty\frac{(\ln(1+x))^k}{(k+1)!}=\sum\limits_{k=0}^\infty\frac{1}{k+1}\sum\limits_{j=k}^\infty (-1)^{j-k}\begin{bmatrix}j\\k\end{bmatrix}\frac{x^j}{j!}=$

$\displaystyle =\sum\limits_{k=0}^\infty\frac{x^k}{k!}\sum\limits_{j=0}^k (-1)^{k-j}\begin{bmatrix}k\\j\end{bmatrix}\frac{1}{j+1}\,$ .