Can we express the following ordinary generating function? I wish to express the following power series 
$$ \sum_{k \ge 0} \binom{n-k}{m} x^k$$
where $n,m$ are positive integer such that  $0< m \le n$
 A: $$
\begin{align}
\sum_{k=0}^\infty\binom{n-k}{m}x^k
&=\sum_{k=0}^n\binom{n-k}{m}x^k+\sum_{k=n+1}^\infty\binom{n-k}{m}x^k\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+\sum_{k=1}^\infty\binom{-k}{m}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^m\sum_{k=1}^\infty\binom{m-1+k}{m}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^m\sum_{k=1}^\infty\binom{m-1+k}{k-1}x^{k+n}\\
&=\sum_{k=0}^n\binom{n-k}{m}x^k+(-1)^mx^{n+1}\sum_{k=1}^\infty(-1)^{k-1}\binom{-m-1}{k-1}x^{k-1}\\
&=\underbrace{\sum_{k=0}^n\binom{n-k}{m}x^k}_{\text{polynomial}}-\frac{x^{n+1}}{(x-1)^{m+1}}
\end{align}
$$
If $n\lt m$, the polynomial part is $0$.
If $n\ge m$, the polynomial part has degree $n-m$:
$$
\begin{align}
\sum_{k=0}^n\binom{n-k}{m}x^k
&=\sum_{k=0}^{n-m}\binom{n-k}{m}x^k\\
&=\sum_{k=m}^n\binom{k}{m}x^{n-k}
\end{align}
$$
A: In fact, the polynomial part $$\sum_{k=0}^n \binom{n-k}{m} x^k= \sum_{k=0}^{n-m} \binom{n-k}{m} x^k + \sum_{k=n-m+1}^n \binom{n-k}{m} x^k$$ 
$$=\sum_{k=0}^{n-m} \binom{n-k}{m} x^k$$
$$ =x^{n} \sum_{k=0}^{n-m} \binom{n-k}{m} x^{k-n} $$ $$=x^{n-m} [\sum_{i=0}^{\infty} \binom{m+i}{m} \frac{1}{x^i} - \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}]$$ 
$$ =x^{n-m} \frac{1}{(1- (1/x))^{m+1}} - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i} $$
$$ = \frac{x^{n+1}}{(x- 1)^{m+1}} - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}.$$
Hence,  $$\sum_{k=0}^{\infty} \binom{n-k}{m} x^k = - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}.$$
