Finding a geometric series for $\frac{1}{(1 + x)^2}$ I'm asked to find a geometric series for $f(x) = \frac{1}{(1 + x)^2}$, I integrate it first and I get $F(x) = \int{\frac{1}{(1 + x)^2}dx} = \frac{1}{-2(1 - (-x))}$
Since $\frac1{1-x} = \sum_{n=0}^\infty x^n$, $F(x) = \frac{-1}2\sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{2}x^n$
If I derive that, $f(x) = \sum_{n=0}^\infty\frac{(-1)^{n+1} n x^{n-1}}{2} = \sum_{n=1}^\infty\frac{(-1)^n (n + 1) x^n}{2}$
But my textbook says that $f(x) = \sum_{n=0}^\infty (-1)^n (n + 1) x^n$
Obviously those 2 things are pretty different (notice the index). What did I get wrong?

HINT
The geometric series is given by
\begin{align*}
\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \ldots = \sum_{n=0}^{\infty}x^{n}
\end{align*}
