Solution to curious infinite series How exactly does one find a closed form to:
$$ \sum_{i=0}^{\infty}\left[\frac{1}{i!}\left(\frac{e^2 -1}{2} \right)^i \prod_{j=0}^{i}(x-2j) \right]$$
When expanded it takes on the form
$$1 + \frac{e^2-1}{2}x + \frac{1}{2!} \left(\frac{e^2-1}{2} \right)^2x(x-2) + \frac{1}{3!} \left(\frac{e^2-1}{2} \right)^3x(x-2)(x-4)... $$
This doesn't appear to be similar to any type of Taylor Series I have seen before. 

HINT:
Use binom expansion:
$$(1+a)^{y}=1+a.y+\frac{a^2}{2!}.y(y-1)+\frac{a^3}{3!}.y(y-1)(y-2)+....$$
$y=\frac{x}{2}$
$$(1+a)^{\frac{x}{2}}=1+a.\frac{x}{2}+\frac{a^2}{2!}.\frac{x}{2}(\frac{x}{2}-1)+\frac{a^3}{3!}.\frac{x}{2}(\frac{x}{2}-1)(\frac{x}{2}-2)+....$$
$$(1+a)^{\frac{x}{2}}=1+\frac{a}{2}.x+\frac{a^2}{2^22!}.x(x-2)+\frac{a^3}{2^33!}.x(x-2)(x-4)+....$$
Then compare with your series
$$1 + \frac{e^2-1}{2}x + \frac{1}{2!} \left(\frac{e^2-1}{2} \right)^2x(x-2) + \frac{1}{3!} \left(\frac{e^2-1}{2} \right)^3x(x-2)(x-4)... $$
Did you see the value of $a$?
