Expansion of $(1+x)^{-n}$ when x is not less than 1. I was told that binomial expansion of $(1+x)^{-n}=1-\dfrac{n}{1!}(x)+\dfrac{n(n+1)}{2!}x^2\cdots$ is only valid when $|x|\lt1$. But what happens when $|x|$ is greater than 1  ($|x|\geq1$) ?
Why it is not defined for x greater than or equal to 1?
I could only think of when x=1,n=1
1/2=1-1+1-1+1...
Which is also 0,1
 A: For $1\le |x|$ the series does not converge.  It is valid only as formal series.
That's because the function $1/(1+x)^n$ has a pole (vertical asymptote) at $x=-1$ and no polynomial in $x$ can represent it on the whole range.
When $1 \le |x|$, you can rewrite it as:
$$
\left( {1 + x} \right)^{\, - n}  = \left( {1 + \left\lfloor x \right\rfloor  + \left\{ x \right\}} \right)^{\, - n}  = \left( {1 + \left\lfloor x \right\rfloor } \right)^{\, - n} \left( {1 + {{\left\{ x \right\}} \over {1 + \left\lfloor x \right\rfloor }}} \right)^{\, - n} \quad \left| {\; - 1 \ne \left\lfloor x \right\rfloor } \right.
$$
where $\left\lfloor x \right\rfloor $ is the floor function and $ \left\{ x \right\}$ is the fractional part
or as:
$$
\left( {1 + x} \right)^{\, - n}  = x^{\, - n} \left( {1 + {1 \over x}} \right)^{\, - n} \quad \left| {\; - 1,0 \ne x} \right.
$$
That is, you are replacing the Taylor series at $x=0$ with a series around ,e.g., $1+\left\lfloor x \right\rfloor $ or with the Laurent series for $1 < |x|< \infty$ (Taylor series for $f(1/x)$ at $1/x \to 0$).
A: $$
\begin{align}
(1+x)^{-n}
&=\frac1{x^n}\left(1+\frac1x\right)^{-n}\\
&=\frac1{x^n}\sum_{k=0}^\infty\binom{-n}{k}\frac1{x^k}\\
&=\frac1{x^n}\sum_{k=0}^\infty(-1)^k\binom{k+n-1}{k}\frac1{x^k}\\
&=\sum_{k=n}^\infty(-1)^{k-n}\binom{k-1}{n-1}\frac1{x^k}\\
\end{align}
$$
