Find general term for $a_{2n}=-(2n-1)(2n-2)a_{2n-2}+2(-1)^{n-1}(2n-2)!$

Solution 1:

Let's solve $$a_{2n}=-(2n-1)(2n-2)a_{2(n-1)}+2(-1)^{n-1}(2n-2)!$$ with $a_2=y^{(2)}(0)=2$.

Define $$b_n = (-1)^{n+1}\frac{a_{2n}}{2\,(2n-1)!}$$ so $$a_{2n} = (-1)^{n+1}2\,(2n-1)!\,b_n$$

Substituting in the recurrence relation for $a$ we have $$(-1)^{n+1}2\,(2n-1)!\,b_n =-(2n-1)(2n-2)(-1)^n2\,(2n-3)!\, b_{n-1} + 2(-1)^{n-1}(2n-2)!$$

Dividing by $(-1)^{n+1}2\,(2n-1)!$ we get

$$b_n = b_{n-1} + \frac{1}{2n-1}$$

This, together with $b_1=1$ gives $$b_n = \sum_{k=1}^n \frac{1}{2k-1}$$ Notice that this sum doesn't have a closed form.

So we end with $$a_{2n}=(-1)^{n+1}2(2n-1)!\sum_{k=1}^n\frac{1}{2k-1}$$