# Negative Binomial Series

On this webpage, the expansion of the negative binomial series is given below.

$$(x+a)^{-n} = \sum^\infty_{k=0} (-1)^k \begin{pmatrix}n + k - 1\\k\end{pmatrix} x^ka^{-n-k}$$ when $$|x| < a$$. My questions are as follows:

1. Are we assuming that $$a > 0$$?
2. What happens when $$|x| < a$$ condition is not true?

I'm not an expert, but I think this condition is not meant to be restrictive, but rather an indication of how to chose $$x$$ and $$a$$ between the two. If $$x = a$$, there is no need for the formula, because $$(x + x)^{-n} = 2^{-n} x^{-n}$$. If both $$x$$ and $$a$$ are negative, you can factor a $$(-1)^{-n} = (-1)^n$$ out of the expression and multiply it in the formula. If one of the terms is positive and the other is negative, check which term has a smaller absolute valua and you can factor $$(-1)^n$$ if the negative one does (Note that the terms can not be the opposite of each other). Then, you choose it to be $$x$$. Otherwise, if both terms are positive, chose the smaller one as $$x$$.

Examples:

Both negative: $$(-1 - 2)^{-n} = (-1)^{n}(1+2)^n$$, $$x=1, a = 2$$

Both positive: $$(3 + 5)^{-n}$$, $$x=3, a=5$$

One positive, one negative: $$(-1 + 3)^{-n}, |-1| < |3|, x=-1, a=3$$, $$(4 - 2)^{-n}, |4| > |-2|, x=-2, a=4$$, $$(-4+2)^{-n} = (-1)^n (4-2)^{-n}$$

Having $$a<0$$ is fine.

The sum on the RHS, as you've written it, doesn't actually converge unless $$\lvert\frac{x}{a}\rvert<1$$. This is why assuming $$|x|<|a|$$ is necessary for using the series expansion.