Validity of binomial expansion for any power

Question

I know that $$(1+x)^n=1+nx+\frac {n(n-1)}{2!}x^2+\frac {n(n-1)(n-2)}{3!}x^3+\frac {n(n-1)(n-2)(n-3)}{4!}x^2+\\ \dots+\frac {n(n-1)\dots(n-r+1)}{r!}x^r\text.$$ I know that this formula is valid for any real number n, provided that $|x|<1$. The formula extends infinitely for a negative value of $n$, and terminates for a positive value of $n$. Is there a formula for a case where $|x|>1$? Why is this formula applicable only for $|x|<1$?

Answer 1

When $x<-1$ note that $(1+x)^{n}$ is not defined for all real $n$. For $x>1$ the series on the right is not convergent. However, you can get an expansion for $x>1$ using the fact that $(1+x)^{n}=x^{n} (1+\frac 1 x)^{n}$ and using the expansion of $(1+\frac 1 x)^{n}$ (which is valid sinec $|\frac 1 x| <1$ in this case).

Answer 2

The series you have written is valid for any real value of $x$ if $n$ is zero or a natural number. In this case your expansion is an identity and you get a polynomial of degree $n+1$ in $x$.

But if $n$ is negative integer or non integral, your expansion is valid only for |x|<1 and the series is infinite and convergent.

For example $$(1+x)^4=1+4x+6x^2+4x^3+x^4.$$ $$(1+x)^{4.3}=1+4.3x+(4.3*3.3)x^2/2+(4.3*3.3*3.2)x^3/6+........$$ $$(1+x)^{-4.3}=1-4.3x+(4.3*5.3)x^2/2-(4.3*5.3*6.3)x^3/6+........$$ $$(1-x)^{-1}=1+x+x^2+x^3+......., \text{if}~ |x|<1$$ $$(1+1/3)^{1/2}=1+\frac{1}{2}(1/3)-\frac{1.1}{2.2.2!}(1/3)^2+\frac{1.1.3}{2.2.2.3!}(1/3)^3+.....$$