A problem about the harmonic oscillator in Quantum mechanics When I learned quantum mechanics by reading Griffith's book called Introduction to quantum mechanics, I was confused by his description.
In Page 53 of the 2ed edition book, after got the recursion formula $a_{j+2}=\frac{(2j+1-K)}{(j+1)(j+2)}a_j$(First we solved the Schr$\ddot o$dinger equation $\frac{d^2\psi}{d\xi^2}=(\xi^2-K)\psi$, where $K=\frac{2E}{\hbar \omega}$, and the solution is that $\psi(\xi)=h(\xi)e^{-\xi^2/2}$. Then we expressed $h(\xi)$ in the form of power series in $\xi$, e.i $h(\xi)=\sum_{j=0}^\infty a_j\xi^j$.) the author took a approximation at very large $j$, the recursion formula becomes $a_{j+2}\approx\frac{2}{j}a_j$. Up to now, I understand everything. But the next things stuck me.
The author's words as follows,

For at very large $j$, the recursion formula becomes (approximately)
  $$
a_{j+2}\approx\frac{2}{j}a_j  \tag{1}
$$
  with the (approximate) solution
  $$
a_j\approx\frac{C}{(j/2)!},\tag{2}
$$
  for some constant $C$, and this yields (at large $\xi$, where the higher powers dominate)
  $$
h(\xi)\approx C\sum \frac{1}{(j/2)!}\xi^j \approx C\sum\frac{1}{j!}\xi^{2j}\approx Ce^{\xi^2} \tag{3}
$$

The question is that I barely understand that  how to deduce equations $(2)$ and $(3)$. In particular, I think the equation $(2)$ is bizarre. And I know if I get $(2)$, I will know how to deduce $(3)$. I hope someone could help me and explain that how to deduce $(2)$. 
 A: You're trying to solve for a series of the sum
$$h(x) = \sum_{j=0}^\infty a_j x^j $$
and you found that only even terms contribute, with
$$a_{j+2} = \text{something } a_j.$$
Since only even terms contribute, let's make our lives easier and rescale the coefficients as follows: we define $b_n \equiv a_{n/2}$ such that
$$h(x) = \sum_{n=0}^\infty b_n x^{2n}$$
Then for very large $n$ you have
$$b_{n+1} \simeq \frac{1}{n+1} b_n + O(1/n).$$
If you forget about the $1/n$ corrections for now, the above recursion is solved exactly for 
$$b_n = \frac{C}{n!}$$
where $C$ is some constant. But that's precisely what you need to show.
A: I see that your problem is you can't deduce equation $(2)$.But note that Hermit polynomial is $h(\xi)=\sum_{j=0}^\infty a_j\xi^j$,and the sum start from 0.So the terms are all even.
Then using
$$a_{j+2}\approx\frac{2}{j}a_j$$
we have$$\begin{align}a_j&\approx\frac{2}{j-2}a_{j-2}\\
&\approx\frac{2}{(j-2)(j-4)}a_{j-4}\\
&\approx\frac{1}{(j/2-1)(j/2-2)(j/2-3)}a_{j-6}\\
&\dots\\
&\approx\frac{a_0}{(\frac{j}{2})!}
\end{align}$$
where $a_0$ is $C$ in your expression.
The rest I'm sure is easy for you to complete.Just put $a_j$ back to $h(\xi)$ and make a variable substitution.
