Newton's (Generalized) Binomial Theorem states that if $r\in\mathbb{R}$ then
$\displaystyle(x+1)^r=\sum_{i=0}^\infty\binom{r}{i}x^i,\quad\mathrm{for}\;x\in[0,1)$.
The series on the right has a radius of convergence of 1. Therefore it converges even if $x\in(-1,0)$. I was wondering what it converges to. Is it still $(x+1)^r$? The proof I know of Newton's Binomial Theorem breaks down if $x\in(-1,0)$.
EDIT: Here is my attempt to proving the formula. I first proved that the radius of convergence of the series is 1 and then I proved that the Maclaurin series of the function $(x+1)^r$ is the series on the right. Lastly, I want to show that the series converges to the function by showing that the remainder of the $i$th Taylor polynomial, in its Lagrange form, tends to zero for $x\in(-1,1)$. This remainder is of the form
$\displaystyle\frac{r(r-1)\cdots(r-i)}{(i+1)!}(c+1)^{r-i-1}x^{i+1}$,
where $c$ is a number between $x$ and $0$.
The absolute value of this quantity is less then
$\displaystyle\frac{|r|^{i+1}}{(i+1)!}|c+1|^{r-i-1}|x|^{i+1}$.
The first and third factor in the product on the right tend to zero as $i$ tends to infinity. But in order for the second factor to tend to zero I need $c$ to be nonnegative, which is guaranteed if we restrict $x$ to the interval $[0,1)$.
Is there a way to fix this proof to make it work for $x\in(-1,0)$?