I posted a question earlier about the taylor of $(1-x)\ln(1-x)$ but i made a miscalculation and decided to delete it, sorry about that.
anyways, i solved the miscalculation and i found that $(1-x)\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}-\sum_{n=1}^{\infty}\frac{x^n}{n}$
Now I just need to combine these two sums into a single sum, and get something of the form $\sum_{n=1}^{\infty}a_nx^n$ and I can't seem to do it without something straggling outside of the sum.
how do i proceed from here?
Edit: we know $\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$, so $\ln(1-x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(-1)^nx^n}{n}=\sum_{n=1}^{\infty}\frac{-x^n}{n}$
and so $(1-x)\ln(1-x)=(1-x)\sum_{n=1}^{\infty}\frac{-x^n}{n}=(x-1)\sum_{n=1}^{\infty}\frac{x^n}{n}=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}-\sum_{n=1}^{\infty}\frac{x^n}{n}$
What only became clear from the comments, it that you want to present it as a Taylor series. This is almost trivial:
$$\begin{align} \sum_{n=1}^{\infty}\frac{x^{n+1}}{n}-\sum_{n=1}^{\infty}\frac{x^n}{n} &= \sum_{n=2}^\infty \frac{x^n}{n-1} - \sum_{n=1}^{\infty}\frac{x^n}{n} \tag 1\\ &= -x + \sum_{n=2}^\infty \left(\frac{1}{n-1} - \frac{1}{n}\right)x^n \tag 2\\ &= \sum_{n=0}^\infty a_n x^n \\ \end{align}$$ with $$ a_n=\begin{cases} 0, & \text { if } n=0 \\ -1, & \text { if } n=1 \\ \dfrac{1}{n-1} - \dfrac{1}{n} = \dfrac1{n(n-1)}, &\text{ if } n\geqslant 2\\ \end{cases}$$
So that $(2)$ is clearly a Tylor series (around 0 with radius of convergence of 1).
The sum $(2)$ converges because the two sums of $(1)$ do converge.