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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.

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    $\begingroup$ Your expansion doesn't match the term in the series, it is for $\prod\limits_{j=0}^{i-1}$, not for $\prod\limits_{j=0}^i$. Which one is the intended? $\endgroup$ – Daniel Fischer Jul 8 '14 at 20:10
  • $\begingroup$ I think the former, since that one actually matches up with the latter series terms correct? $\endgroup$ – frogeyedpeas Jul 8 '14 at 20:13
  • $\begingroup$ Well, I don't know what the intention is. However, if the product goes only to $i-1$, you get a nicer result, $e^x$, so that makes it sort of likely. $\endgroup$ – Daniel Fischer Jul 8 '14 at 20:16
  • $\begingroup$ Yes that seems correct, but the logic is still the same as your response? $\endgroup$ – frogeyedpeas Jul 8 '14 at 20:17
  • $\begingroup$ Yes, the logic is the same. Note that with $i+1$ factors in the product, I had to pull out the one from the sum. With only $i$ factors, it is even nicer. $\endgroup$ – Daniel Fischer Jul 8 '14 at 20:18
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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 $a$?

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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

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

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    $\begingroup$ Your idea is great and awesome! :) $\endgroup$ – Mohammad Khosravi Jul 8 '14 at 20:10

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