Negative Binomial Series

Question

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?

Answer 1

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

Answer 2

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.