Solution to curious infinite series

Question

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.

Answer 1

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 $a$?

Answer 2

Note that

$$\frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^i (x-2j) = x \prod_{j=1}^i \frac{\frac{x}{2}-j}{j} (e^2-1)^i = x\binom{\frac{x}{2}-1}{i}(e^2-1)^i,$$

so

$$\sum_{i=0}^\infty \frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^i (x-2j) = x\sum_{i=0}^\infty \binom{\frac{x}{2}-1}{i}(e^2-1)^i = x\left(1+(e^2-1)\right)^{\frac{x}{2}-1} = x e^{x-2}.$$

The clue is to recognise the binomial coefficient in $$\frac{1}{i!}\prod_{j=0}^i(x-2j).$$

If, as the expanded form indicates, the product is $\prod\limits_{j=0}^{i-1} (x-2j)$ instead of $\prod\limits_{j=0}^{i} (x-2j)$, we have

$$\frac{1}{i!}\prod_{j=0}^{i-1}\left(\frac{x}{2}-j\right) = \prod_{k=1}^i \frac{\frac{x}{2}+1-k}{k} = \binom{\frac{x}{2}}{i},$$

and

$$\sum_{i=0}^\infty \frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^{i-1} (x-2j) = \sum_{i=0}^\infty \binom{\frac{x}{2}}{i}(e^2-1)^i = \bigl(1+(e^2-1)\bigr)^{\frac{x}{2}} = e^x.$$