power series with binomial coefficient

Good evening, I don't know how to find the function associate with this power series : $$\sum_{n=k}^\infty \binom{n}{k} x^n$$

Should I look for a differential equation ?

Thx in advance

Solution 1:

\begin{align} & \sum_{n=k}^\infty \frac{n(n-1)(n-2)\cdots (n-k+1)}{k!} x^{n-k}x^k \\[10pt] ={} & \frac{x^k}{k!} \sum_{n=k}^\infty \frac{d^k}{dx^k} x^n \\ & \text{Pulling out $x^k/k!$ works because $k$ does not change as $n$ changes.} \\[10pt] = {} & \frac{x^k}{k!} \, \frac{d^k}{dx^k} \sum_{n=0}^\infty x^n \\ & \text{This works because power series can be differentiated} \\ & \text{term by term. Here we need not start at $k$ since the} \\ & \text{derivatives of the first $k-1$ terms are $0.$} \\[10pt] = {} & \frac{x^k}{k!} \, \frac {d^k}{dx^k} \, \frac 1 {1-x} = \frac{x^k}{k!} \, k!(1-x)^{-(k+1)} = \frac{x^k}{(1-x)^{k+1}}. \end{align}