I came across the following binomial identity which I had never previously encountered.
$$(1+x)^n = \left(\frac{1}{1+x}\right)^{-n} = \left[\frac{1}{2}\left(1 + \frac{1-x}{1+x}\right) \right]^{-n} = 2^n\left(1 + \frac{1-x}{1+x}\right)^{-n}$$
which, given the negative exponent, can be expanded as $$ 2^n\left(1 - n\left(\frac{1-x}{1+x}\right) + \frac{n(n+1)}{2!}\left(\frac{1-x}{1+x}\right)^2 + \dots \right) = 2^n\sum_{k=0}^{\infty} (-1)^k \frac{(-n)_k}{k!}\left(\frac{(1-x)}{(1+x)}\right)^k $$
I'm aware of the binomial series for fractional and negative exponents. But typically a binomial series for $n \in \mathbb{Z}^+$ is described as, sort of, "collapsing" into the standard (finite) binomial formula, (since eventually terms have an $n - n$ factor and thus go to $0$). This lead me, perhaps naively, to believe there was no way of writing $(1 +x)^n$ as an infinite series when $n \in \mathbb{Z}^+$
Restrictions on $x$ and convergence notwithstanding, does this look correct? If so, is this an identity of any significant utility or just a curiosity?
This will converge where $|\frac{1-x}{1+x}| < 1$, which holds for all $x > 0$.
This is similar to the combining of $\ln(1-x) =-x-\frac{x^2}{2}-\frac{x^3}{3}-...$ with $\ln(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}-...$ to get $\ln(\frac{1+x}{1-x}) =2(x+\frac{x^3}{3}+\frac{x^5}{5}+...) $.
From this, if $y=\frac{1+x}{1-x}$ then $x=\frac{y-1}{y+1} $ and given a positive $y$, the series $\ln(y) =\ln(\frac{1+x}{1-x}) =2(x+\frac{x^3}{3}+\frac{x^5}{5}+...) $ will converge.