Series expansion of $ \frac{x}{\ln (1+x)}$

Question

What are coefficients in the expansion of series for

$$ \frac{x}{\ln (1+x)} = \sum_{n=0}^\infty A_n \frac{x^n}{n!}?$$

Do they have a name?

Answer 1

It is known, that $$ \frac{z}{\ln(1-z)} = \sum_{n=0}^{\infty}C_nz^n, \quad |z|<1, $$ where the coefficients, $C_n$ are called Gregory coefficients. They can by calculated by the following recursion for all $n \geq 0$. From equations: $$ C_0 = -1,\quad \sum_{k=0}^n\frac{C_k}{n+1-k} = 0,\quad n=1,2,3,\ldots $$ If we use the $z=-x$ substituion, we get $$ -\frac{x}{\ln(1+x)} = \sum_{n=0}^{\infty}(-1)^nC_nx^n, \quad |x|<1. $$ The first few Gregory coefficients are: $$ -1,\frac12, \frac{1}{12}, \frac{1}{24}, \frac{19}{720},\frac{3}{160},\frac{863}{60480},\frac{275}{24192},\frac{33953}{3628800},\dots $$

Answer 2

One may observe that, by the binomial theorem, $$ (1+x)^t=1+\sum _{n=1}^{\infty } \frac{\prod _{k=0}^{n-1} (t-k)}{n!}x^n, \quad|x|<1, \tag1 $$ then, integrating with respect to $t$ from $t=0$ to $t=1$, one gets $$ \frac{x}{\ln(1+x)}=\sum_{n=0}^\infty A_n \frac{x^n}{n!} $$ with

$$ A_0=1,\quad A_n=\int_0^1 t(t-1)(t-2)\cdots(t-n+1)\:dt, \quad n \geq1. $$