Simplify the series given by the recurrence relation $na_n=2a_{n-2}$

Question

If you are given a recurrence relation such that:

$$na_n=2a_{n-2}\implies a_n= \begin{cases} 0 & \text{odd} \,n \\ \frac{2}{n}a_{n-2} & \text{even} \,n \end{cases}$$

My textbook suggests that the series can be simplified by

Putting $n=2m$ (since only even terms appear in this series), we get $$a_{2m}=\frac{2}{2m}a_{2m-2}=\bbox[#AF0]{\frac{1}{m}a_{2m-\color{red}{2}}=\frac{1}{m}\color{red}{\frac{1}{(m-1)}}a_{2m-\color{red}{4}}}=\frac{1}{m!}a_0$$

I understand the final equality as $$\frac{1}{m}\frac{1}{(m-1)}\frac{1}{(m-2)}\frac{1}{(m-3)}\cdots=\frac{1}{m!}$$

But I do not understand the highlighted equality.

Why does the $\color{red}{\cfrac{1}{m-1}}$ appear when $a_{2m-\color{red}{2}}$ is reduced to $a_{2m-\color{red}{4}}$?

Answer 1

From $$ a_n=\frac{2}{n}\:a_{n-2},\quad n\,\, \text{even} $$ you may just put $n=2m-2$ to obtain $$ a_{2m-\color{red}{2}}=\color{blue}{\frac{2}{2m-2}}\:a_{2m-\color{red}{2-2}} $$ or $$ a_{2m-\color{red}{2}}=\color{blue}{\frac1{m-1}}\:a_{2m-\color{red}{4}}. $$

Answer 2

What has been provided is $$a_{2m} = \frac{1}{m} \, a_{2m-2}$$ From this, then \begin{align} a_{2m} &= \frac{1}{m} \, a_{2(m-1)} \\ &= \frac{1}{m \, (m-1)} \, a_{2(m-2)} \\ &= \frac{1}{m \, (m-1) \, (m-2)} \, a_{2(m-3)} ... \end{align} The pattern is $$a_{2m} = \frac{(m-k)!}{m!} \, a_{2(m-k)},$$ where $0 \leq k \leq m$. When $k=m$ then $a_{2m} = \frac{a_{0}}{m!}$.