Use Differentiability of Power Series to find the sum $F(x) = \sum_{n=0}^{\infty} \frac{(x+1)^{n+1}}{n+1}$

Question

$$F(x) = \sum_{n=0}^{\infty} \frac{(x+1)^{n+1}}{n+1}$$

I found the following values after differentiating:

$$ (x+1)^n, n(x+1)^{n-1}, n(n-1)(x+1)^{n-2} $$

It looks a lot like the Sum for $F(x)= 1/(1-x)$, but i'm not sure where to proceed from here. Thanks for your help.

Answer 1

With differentiating of $$f(x)=\sum_{n\geq0}\dfrac{(x+1)^{n+1}}{n+1}$$ we have $$f'(x)=\sum_{n\geq0}(x+1)^n=\dfrac{1}{1-(x+1)}=-\dfrac1x$$ valid for $|x+1|<1$, then with integration of both sides $$f(x)=\int_{-1}^x\dfrac{1}{-t}\ dt=-\ln(-x)$$