I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$:
$${n \choose k} = (-1)^k {-n + k – 1 \choose k}$$
So now let's look at one case for using the binomial coefficient:
$$(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$$
How do I evaluate $\sum_{k = 0}^{n}$ when $n < 0$? From searching around on the internet I think it's just an infinite series, i.e. $k$ keeps incrementing by 1 forever. But that gets me confused about
$$\begin{align*}
(a + b)^n &= a^n(1 + \frac{b}{a})^n \\ &= a^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{b}{a}\right)^k\right)\\
&= a^n \left(1 + n \left(\frac{b}{a}\right) + \frac{(n)(n-1)}{2}\left(\frac{b}{a}\right)^2 + \cdots\right)
\end{align*}$$
and
$$\begin{align*}
(b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\
&= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\
&= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right)
\end{align*}$$
Now the two should be equal, but in the first sum I'd never get a $b^n$ and in the second sum I'd never get a $a^n$?
Best Answer
The below is too long for a comment so I'm including it here even though I'm not sure it "answers" the question.
If you think about $(1+x)^{-n}$ as living in the ring of formal power series $\mathbb{Z}[[x]]$, then you can show that $$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$ and the identity $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ seems very natural. Here's how...
First expand $(1+x)^{-n} = \bigg(\frac{1}{1-(-x)}\bigg)^n = (1 - x + x^2 - x^3 + \dots)^n$. Now, the coefficient on $x^k$ in that product is simply the number of ways to write $k$ as a sum of $n$ nonnegative numbers. That set of sums is in bijection to the set of diagrams with $k$ stars with $n-1$ bars among them. (For example, suppose $k=9$ and $n=4$. Then,
**|*|***|***corresponds to the sum $9=2+1+3+3$;****||***|**corresponds to the sum $9 = 4+0+3+2$;****|***||**corresponds to $9=4+3+0+2$; etc.) In each of these stars-and-bars diagrams we have $n+k-1$ objects, and we choose which ones are the $k$ stars in $\binom{n+k-1}{k}$ many ways. The $(-1)^k$ term comes from the alternating signs, and that proves the sum.