How to derive a generating function when given a sequence

I am asked to derive a generating function for the sequence

$2,2\sqrt{2}, \sqrt{2}(\sqrt{2} - 1), \frac13\sqrt{2}(\sqrt{2}-1)(\sqrt{2}-2), \frac{1}{(3)(4)}\sqrt{2}(\sqrt{2}-1)(\sqrt{2}-2)(\sqrt{2}-3), \frac{1}{(3)(4)(5)}\sqrt{2}(\sqrt{2}-1)(\sqrt{2}-2)(\sqrt{2}-3)(\sqrt{2}-4),...$

I have no idea how to approach this problem. Usually when I'm asked to derive a generating function for a sequence I start with a familiar generating function and sequence such as $\frac{1}{1-x}$ and then manipulate it till it matches the given sequence. For this sequence however, I'm only recognizing the pattern of $\sqrt{2}!$ and not sure where to continue from there. Any help with the problem or a better strategy for deriving generating functions is appreciated.

You can notice that the terms have a nice-looking ratio

$$\frac{a_n}{a_{n-1}} = \frac{\sqrt{2} - (n-1)}{n}$$

which doesn't immediately help with standard generating function techniques, but you might recognise that it looks like a binomial coefficient:

$$\binom{\alpha}{n} \div \binom{\alpha}{n-1} = \frac{\alpha - (n - 1)}{n}$$

And so each term $a_n$ in your sequence is actually $a_n = 2 \binom{\sqrt{2}}{n}$. Hence the generating function is

\begin{aligned} f(x) &= 2 \sum_{n \geq 0} \binom{\sqrt{2}}{n} x^n \\ &= 2 (1 + x)^\sqrt{2} \end{aligned}

I don't know a better way of coming up with this than already being familiar with binomial coefficients.

• Very nice work. – marty cohen Sep 27 '17 at 5:31

You're a life saver. These question are just so damn intimidating, especially since we've just started the topic.

• This is a comment, not an answer to question. – Lord Shark the Unknown Sep 27 '17 at 6:42