For what value of $x$ does the series $$\sum_{}^{}\dfrac{(1+x)^n}{n(n-1)}$$ converge?
Show that on a certain range of $x$ it determines a differentiable function whose derivative is $\log(-x)$.
Now my attempt at the answer is as follows;
If we use ratio test; then we get $$\left|\frac{(1+x)^{n+1}}{(n+1)n}\dfrac{n(n-1)}{(1+x)^{n}}\right|=\left|(1+x)\frac{(n-1)}{(n+1)}\right|$$ and for it to converge the limit as $n \rightarrow \infty$ must be $<1$ ,so as $n \rightarrow \infty$ we get $$|x+1|<1$$ and only value that satisfy this are $x \in (-2,0)$.
Now the second part is that which confuses I just can't see the relationship between the derivative of the above power series and $\log(-x)$. Any help would be appreciated.
The derivative of the given sum which is a power series is $$\sum_{n=2}^\infty\frac{(1+x)^{n-1}}{n-1}=\sum_{n=1}^\infty\frac{(1+x)^{n}}{n},\quad\forall x: |x+1|<1$$ but with $z=x+1$ we have $|x+1|<1\iff|z|<1$ and $$\sum_{n=1}^\infty\frac{(1+x)^{n}}{n}=\sum_{n=1}^\infty\frac{z^{n}}{n}=-\log(1-z)=-\log(-x)$$