How prove this $\sum_{n=1}^{\infty}\frac{x^n(1-x)^n}{n!}f^{(n)}(x)=-\frac{1}{2}xf(x)$ let $$f(x)=\dfrac{x}{\ln{(1-x)}}$$
prove that for $0<x<1$,
$$\sum_{n=1}^{\infty}\dfrac{x^n(1-x)^n}{n!}f^{(n)}(x)=-\dfrac{1}{2}xf(x)$$
I think use this  Taylor series
$$f(x)=f(t)+f'(t)(x-t)+f''(t)(x-t)^2/2+\cdots+\dfrac{f^{(n)}(t)}{n!}(x-t)^n+\cdots$$
then I can't.Thank you.
 A: Look at $\;\displaystyle f(z) = \frac{z}{\log(1-z)}\;$ as a function over $\mathbb{C}$. It is analytic over the whole $\mathbb{C}$ except along a branch cut $[1,\infty)$ on the real axis. Even though the factor $\frac{1}{\log(1-z)}$ has a pole at $z = 0$, the pole get cancelled by the numerator $z$ there. 
For every $z \in \mathbb{C}$ such that $\Re z < 1$, the nearest singularity of $f$ over $\mathbb{C}$ is the point $z = 1$. This implies if one take any $t \in \mathbb{C}$ such that $|t| < |z-1|$, the Taylor series expansion:
$$\sum_{n=0}^\infty f^{(n)}(z)\frac{t^n}{n!} \;\;\text{ converges to }\;\; f(z+t)$$
For any $x \in (0,1)$, we have $\Re x < 1$ and $|x(1-x)| < |x-1|$. This implies the desired
sum
$$\sum_{n=1}^\infty f^{(n)}(x)\frac{x^n(1-x)^n}{n!}\;\;\text{ converges to }\;\;f(x+x(1-x)) - f(x)$$
With a little bit of algebra, we obtain:
$$\begin{align}
f(x+x(1-x))-f(x) 
&= \frac{x + x(1-x)}{\log(1 - (x+x(1-x)))} - \frac{x}{\log(1-x)}\\
&= \frac{2x - x^2}{2\log(1-x)} - \frac{x}{\log(1-x)}\\
&= - \frac{x^2}{2\log(1-x)}\\
&= -\frac{x}{2}f(x)
\end{align}$$
