Expansion of $(1+x)^{-n}$ when x is not less than 1.

Question

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

Answer 1

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 $1/x \to 0$).

Answer 2

$$ \begin{align} (1+x)^{-n} &=\frac1{x^n}\left(1+\frac1x\right)^{-n}\\ &=\frac1{x^n}\sum_{k=0}^\infty\binom{-n}{k}\frac1{x^k}\\ &=\frac1{x^n}\sum_{k=0}^\infty(-1)^k\binom{k+n-1}{k}\frac1{x^k}\\ &=\sum_{k=n}^\infty(-1)^{k-n}\binom{k-1}{n-1}\frac1{x^k}\\ \end{align} $$