Computing a finite sum involving binomial coefficients I would like to prove the following: given the sequence $$a_m = \begin{cases}1 & \text{if $m = 0, 1$} \\\frac{(\alpha+2)(\alpha+4)\cdots(\alpha+2(m-1))}{(\alpha+1)(\alpha+2)\cdots(\alpha+m-1)} & \text{otherwise}\end{cases}$$ then, $$\sum_{m=0}^{n}(-1)^m\binom{n}{m}a_m = \begin{cases}\frac{(n-1)!!}{(\alpha+1)(\alpha+3)\cdots(\alpha+n-1)}& \text{if $n$ is even and $\ge 2$} \\0 & \text{if $n$ is odd}\end{cases}$$
I understand that the left hand side in the last equation is the $n$-th difference $\Delta^na_0$, so basically I tried to compute each difference but it seems that a general formula is hard to find for $\Delta^na_k$. Can you help me? I have seen often that such problems can be attacked with special functions but unfortunately I don't have any experience :(. What do you suggest? Any hint would be greatly appreciated :)
EDIT: So here is the original problem: show that $$e^{-x}\left[1+x+\frac{\alpha+2}{\alpha+1}\cdot\frac{x^2}{2!}+\frac{(\alpha+2)(\alpha+4)}{(\alpha+1)(\alpha+2)}\cdot\frac{x^3}{3!}+\ldots\right]=$$$$1+\frac1{\alpha+1}\frac{x^2}{2^1\cdot1!}+\frac1{(\alpha+1)(\alpha+3)}\frac{x^4}{2^2\cdot2!}+\ldots$$. Doing the product one just compares the two series and obtains my equation above (if I interpreted correctly the pattern of the two series).
 A: First of all, you do not need to precise $a_0=a_1=0$ since they are explicit from the formula which gives $a_m$.
Next, the $a_m$ simplify a lot when $m$ is even. For exemple
$$a_7=\frac{(\alpha +8) (\alpha +10) (\alpha +12)}{(\alpha +1) (\alpha +3) (\alpha +5)}\qquad a_8=\frac{(\alpha +8) (\alpha +10) (\alpha +12) (\alpha +14)}{(\alpha +1) (\alpha +3)
   (\alpha +5) (\alpha +7)}$$
Using Pochhammer symbols
$$a_m=2^{m-1}\frac{ \left(\frac{\alpha }{2}+1\right)_{m-1}}{(\alpha +1)_{m-1}}$$
$$S_n=\sum_{m=0}^{n}(-1)^m\binom{n}{m}\,a_m =\frac{1+(-1)^n}{2 \sqrt{\pi } }\,\frac{ \Gamma \left(\frac{\alpha +1}{2}\right) \Gamma
   \left(\frac{n+1}{2}\right)}{ \Gamma \left(\frac{1}{2} (n+\alpha
   +1)\right)}$$ If I am not mistaken, we do not obtain the same formula for $S_n$ when $n$ is even.
A: From the definition, of the Pochhammer symbol, $$(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$$we can express
\begin{equation}
  a_m=2^m\frac{ \left(\frac{\alpha }{2}\right)_{m}}{(\alpha )_{m}}
 \end{equation}
with $a_0=a_1=1$. (The expression given by @ClaudeLeibovici can be retrieved by remarking that
\begin{equation}
  (s+1)_{m-1}=\frac{\Gamma(s+m)}{\Gamma(s+1)}=\frac{1}{s}\frac{\Gamma(s+m)}{\Gamma(s)}=\frac{1}{s}(s)_m
 \end{equation}
The  series proposed in the edit can be expressed as a confluent hypergeometric function
\begin{align}
 \sum_{k=0}^\infty a_k\frac{x^k}{k!}&= 1+x+\frac{\alpha+2}{\alpha+1}\cdot\frac{x^2}{2!}+\frac{(\alpha+2)(\alpha+4)}{(\alpha+1)(\alpha+2)}\cdot\frac{x^3}{3!}+\ldots\\
 &=\sum_{k=0}^\infty\frac{ \left(\frac{\alpha }{2}\right)_{k}}{(\alpha )_{k}}\frac{(2x)^k}{k!}\\
 &=\,_1F_1\left(  \frac{\alpha}{2},\alpha;2x \right)
 \end{align}
It can be recognized as related to a modified Bessel function
\begin{equation}
  I_\nu(z)=\frac{2^{-\nu}z^\nu e^{-z}}{\Gamma(\nu+1)}\,_1F_1\left(  \nu+\frac{1}{2},2\nu+1;2z \right)
 \end{equation}
Here, taking $\nu=(\alpha-1)/2$, we find
\begin{equation}
  \sum_{k=0}^\infty a_k\frac{x^k}{k!}=2^{\frac{\alpha-1}{2}}\Gamma\left( \frac{\alpha+1}{2} \right)x^{\frac{1-\alpha}{2}}e^{x}I_{\frac{\alpha-1}{2}}(x)
 \end{equation}
Then
\begin{equation}
  e^{-x}\sum_{k=0}^\infty a_k\frac{x^k}{k!}=\Gamma\left( \frac{\alpha+1}{2} \right)\left( \frac{x}{2} \right)^{\frac{1-\alpha}{2}}e^{x}I_{\frac{\alpha-1}{2}}(x)
 \end{equation}
From the series expansion
\begin{align}
  I_\nu(z)&=\sum_{k=0}^\infty \frac{1}{\Gamma(k+\nu+1)k!}\left( \frac{z}{2} \right)^{2k+\nu}\\
  &=\frac{1}{\Gamma(\nu+1)}\left( \frac{z}{2} \right)^\nu\left( 1+\frac{z^2}{4(\nu+1)}+\frac{z^4}{32(\nu+1)(\nu+2)}+\ldots \right)
 \end{align}
we obtain
\begin{equation}
   e^{-x}\sum_{k=0}^\infty a_k\frac{x^k}{k!}=1+\frac1{\alpha+1}\frac{x^2}{2^1\cdot1!}+\frac1{(\alpha+1)(\alpha+3)}\frac{x^4}{2^2\cdot2!}+\ldots
 \end{equation}
as expected
