Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum of $\sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$ and concluding that $f(x)=(x-1)\log(1 - 2 x) -2x$ or to replace $\log(1 - 2 x)$ for it's Taylor Expansion and resaulting in $ \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$. I've tried both ways, starting from the Taylor Expansion of $\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n}$ but I've failed both ways. Any hints on how to prove this? Thanks in advance.
Edit: What I did
$$\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n} = -2x + \sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} \\ \rightarrow  (x-1)\log(1 - 2 x)=-2x^2 + x\sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} + 2x +\sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ \rightarrow (x-1)\log(1 - 2 x) -2x =-2x^2 + \sum_{n=2}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^{n} x^{n+1}}{n} + \sum_{n=1}^{\infty} \frac{2^{n+1} x^{n+1}}{n+1} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1} (n+1) + 2^{n+1} x^{n+1} n}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}[(n+1)-2n]}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}(1-n)}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = 0 + \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ =  \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)}$$
Which is what I was trying to prove.
 A: Let $f(x)=(x-1)\log(1-2x)-2x$, then $f$ is differentiable and
$$ f'(x)=(x-1)\frac{-2}{1-2x}+\log(1-2x)-2=\log(1-2x)+\frac{1}{1-2x}-1$$
For $|x|<\frac{1}{2}$, we have
$$ \log(1-2x)=-\sum_{n=1}^{+\infty}\frac{(2x)^n}{n} \text{ and } \frac{1}{1-2x}=\sum_{n=0}^{+\infty}(2x)^n$$
thus
$$ f'(x)=\sum_{n=1}^{+\infty}(2x)^n\frac{n-1}{n}=\sum_{n=2}^{+\infty}(2x)^n\frac{n-1}{n} $$
Now let $g(x)=\sum_{n=2}^{+\infty}\frac{2^n(n-1)x^{n+1}}{n^2+n}$, then $g$ is differentiable and
$$ g'(x)=\sum_{n=2}^{+\infty}2^n\frac{n-1}{n}x^n=f'(x) $$
Thus, since $f(0)=g(0)=0$, we have $f(x)=g(x)$ for $|x|<\frac{1}{2}$.
A: $$\log(1-2x)=-\sum_{n=1}^\infty\frac{(2x)^n}n\;,\;\;|2x|<1\iff |x|<\frac12\implies$$
$$(x-1)\log(1-2x)-2x=\sum_{n=1}^\infty\frac{2^nx^{n+1}-2^nx^n}n-2x$$
and
$$\sum_{n=1}^\infty\frac{2^n(n-1)x^{n+1}}{n^2+n}=\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n+1}}^{:=f(x)}-\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n^2+n}}^{:=g(x)}$$
and now
$$f'(x)=\sum_{n=1}^\infty2^nx^n\;,\;\;g'(x)=\sum_{n=1}^\infty\frac{2^nx^n}n\implies f'(x)-g'(x)=\frac1{1-2x}-1+\log(1-2x)$$
Now try to take it from here...
A: Another approach.
$$f(x)=(x-1)\log(1 - 2 x) -2x \implies f''(x)=\frac{4 x}{(1-2 x)^2}$$ Let $t=2x$ to make
$$g''(t)=\frac{2 t}{(1-t)^2}=2\sum_{n=1}^\infty n t^n\implies g'(t)=2\sum_{n=1}^\infty\frac{ n t^{n+1}}{n+1}\implies g(t)=2\sum_{n=1}^\infty\frac{ n t^{n+2}}{(n+1) (n+2)}$$ Replace $t$ by $2x$ and shift the index by $1$
