<s>[INST] Evaluating an expression using snake oil and convolutions gives different answers I have to evaluate this expression $\sum \limits_{k=0}^n(-1)^k\binom{n}{k}\binom{m+n-k}{n-k}$ using snake oil and convolutions. The problem is that I got two different results, could you help me to find the mistake?
(Notation: $[x^n]$ is the operator "take the coefficient of $x^n$.")
Convolution: 
$$
\begin{eqnarray*}
\sum_{k=0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k}
&=& [x^n] \left( \sum_k(-1)^k \binom{n}{k}x^k \right) \left( \sum\binom{m+k}{k}x^k \right) 
\\ &=& [x^n] (1-x)^n \frac{1}{(1-x)^{m+1}} 
\\ &=& [x^n] (1-x)^{n-m-1}
\\ &=& (-1)^n\binom{n-m-1}{n}
\end{eqnarray*}
$$
Snake oil: 
$$
\begin{eqnarray*}
\sum_m \sum_{k=0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k} x^m 
&=& 
\sum_{k=0}^n (-1)^k \binom{n}{k} \sum_m \binom{m+n-k}{n-k} x^m
\\ &=&
\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{(1-x)^{n-k}}
\\ &=&
\frac{1}{(1-x)^n} \sum_{k=0}^n (-1)^k \binom{n}{k}(1-x)^k
\\ &=&
\frac{x^n}{(1-x)^n}
\\ &=&
\sum_m \binom{m-1}{n-1} x^m ,
\end{eqnarray*}
$$ 
and so the value should be $\binom{m-1}{n-1}$.
Could you help me to find the mistake, please?
 [/INST] I think there's an error in the "snake oil" part. You should have
$$\sum_{m = 0}^{\infty} \binom{m+n-k}{n-k} x^m = \frac{1}{(1-x)^{n+1-k}}$$
Which leads you to
$$\sum_{m = 0}^{\infty} \sum_{k = 0}^n (-1)^k \binom{n}{k} \binom{m+n-k}{n-k} x^m = \frac{x^n}{(1-x)^{n+1}} = \sum_{m = 0}^{\infty} \binom{m}{n} x^m$$
And finally, it's easy to check
$$\binom{m}{n} = (-1)^n \binom{n-m-1}{n}$$
</s>