Question: Find the number of ways that 2n people may be paired.
I have figured this problem out, and I'm fairly certain that there are $\frac{(2n)!}{2^{n} n!}$ ways. However, I cannot seem to work this out using generating functions.
Consider the recurrence relation for this problem. It is given by $a_{n} = (2n-1)a_{n-1}$. It is immediate by inspection that this is just $(2n-1) \cdot (2n-3) \dots 3 \cdot 1$, which agrees with my above answer.
However, this does not agree with the way I have been instructed to solve recurrence relations (other than through the use of generating functions), which is to set $a_{n} = \alpha^{n}$, and solve for $\alpha$, which gives us $a_{n} = (2n-1)^{n}$. Why did this approach not work out?
And as far as solving this using generating functions, I've worked out that $$g(x) - a_{0} = \displaystyle\sum_{n=1}^{\infty}a_{n}x^{n}= \displaystyle\sum_{n=1}^{\infty}(2n-1)a_{n-1}x^{n}= 2\displaystyle\sum_{n=1}^{\infty}n a_{n-1}x^{n} - \displaystyle\sum_{n=1}^{\infty}a_{n-1}x^{n}$$
This reduces to:
$$g(x)-a_{0} = 2x\displaystyle\sum_{m=0}^{\infty}(m+1)a_{m}x^{m} - xg(x)$$
I'm having trouble re-expressing the power series with the $(m+1)$ multiplier. To this end, I would like to ask if there is a straightforward way to re-express a power series of the form $\displaystyle\sum_{m=0}^{\infty}m \cdot a_{m}x^{m}$?
I know that $x\left(\frac{d}{dx} \frac{1}{1-x}\right) = \displaystyle\sum_{m=0}^{\infty}mx^{m}$, so I figure that $x\left(\frac{d}{dx} g(x) \right)$ should do the trick, but I wasn't able to make that approach work out.
Order your people in some way. This can be done in $(2n)!$ ways. then pick the first and second, third and fourth, etc, as your pairing. Obviously switching the first and second person will not change the pairing, so you divide by $2^n.$ Reordering the pairs arbitrarily also does not change the pairing, so that gives you another $n!,$ for a final answer of $(2n)!/(2^n n!).$
If user4140 is correct, and it is generating functions you care about: the method you describe only works for recurrences with constant coefficients, which yours is not. Your method (which gives a perfectly correct answer) is fine.