Prove for integer $k \ge 0$, $\frac1{(1 - x)^{k+1}}= \sum_{i=k}^{\infty}\binom{i}{k}x^{i-k}$

Question

Prove for integer $k \ge 0$, $$\frac1{(1 - x)^{k+1}}= \sum_{i=k}^{\infty}\binom{i}{k}x^{i-k}$$

I've looked up resources for generating functions. However, this specific example seems to be difficult to prove.

Answer 1

You can prove this directly! Begin with the (formal) geometric series $$ \frac{1}{1-x} = \sum_{i=0}^{\infty} x^i. $$ Note that $$ \frac{d}{dx} \left[ \frac{1}{1-x} \right]= \frac{1}{(1-x)^2}, $$ and by induction,
$$ \frac{d^k}{dx^k} \left[ \frac{1}{1-x} \right] = \frac{k!}{(1-x)^{k+1}}, $$ for any $k \geq 0$.

So it suffices to calculate the $k$th derivative of the geometric series. Note that for all orders $0 \leq i < k$, $$ \frac{d^k}{dx^k} \left[ x^i \right] = 0, $$ and for $i \geq k$ $$ \frac{d^k}{dx^k} \left[ x^i \right] = \frac{k \cdot (k-1) \cdots (k-i+1)}{i \cdot (i-1) \cdots 1} \, x^{i-k} = \frac{k!}{(i-k)!} \, x^{i-k}. $$ Thus, \begin{align} \frac{1}{(1-x)^{k+1}} &= \frac{1}{k!} \frac{d^k}{dx^k} \left[ \frac{1}{1-x} \right] \\[2pt] &= \frac{1}{k!} \frac{d^k}{dx^k} \left[ \sum_{i=0}^{\infty} x^i \right] \\[2pt] &= \frac{1}{k!} \frac{d^k}{dx^k} \left[ \sum_{i=k}^{\infty} x^i \right] \\[2pt] &= \sum_{i=k}^{\infty} \frac{1}{k!} \frac{d^k}{dx^k} \left[ x^i \right] \\[2pt] &= \sum_{i=k}^{\infty} \frac{i!}{k!\,(i-k)!} \, x^{i-k} \\[2pt] &= \sum_{i=k}^{\infty} \binom{i}{k} \, x^{i-k}, \end{align} as desired.