Transforming power series Definition:
Let $$a_0 = a_1 = 1, \; a_{n+2} = a_{n+1} + (n+1) \cdot a_n, \; n \geq 0$$
Exercise:
Prove that $$\sum_{n\geq 0} \frac{a_n}{n!} x^n = \exp \left( x + \frac{1}{2} x^2 \right)$$

I don't know how to start with this. I do know that $$e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n, \; e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ but how do I get this wrapped up?
Thank you in advance!
 A: HINT $\ $ The differential equation for the exponential yields a recurrence for the coefficients, for example see here and here.
A: Here's a probabilistic interpretation/derivation of the result. Henceforth, we assume that $x \neq 0$ (the case $x=0$ being trivial).
Recall that the moment-generating function of $X \sim {\rm N}(\mu,\sigma^2)$ is given by
$$
{\rm E}[e^{tX} ] = e^{\mu t + \sigma ^2 t^2 /2} .
$$
Thus setting $t=1$, $\mu=x$, and $\sigma^2 = x^2$, we get
$$
{\rm E}[e^{X} ] = e^{x + x^2 /2}.
$$
On the other hand,
$$
{\rm E}[e^{X} ] = \sum\limits_{n = 0}^\infty  {\frac{{\mu _n }}{{n!}}} ,
$$
where $\mu_n = {\rm E}(X^n)$ is the $n$th moment of the ${\rm N}(x,x^2)$ distribution.
Using integration by parts, we find
$$
\mu_n = \frac{1}{{\sqrt {2\pi x^2 } }}\int_{ - \infty }^\infty  {u^n \exp \bigg[ - \frac{{(u - x)^2 }}{{2x^2 }}\bigg]du}  = \frac{1}{{\sqrt {2\pi x^2 } }}\int_{ - \infty }^\infty  {\frac{{u^{n + 1} }}{{n + 1}}\exp \bigg[ - \frac{{(u - x)^2 }}{{2x^2 }}\bigg]\frac{{u - x}}{{x^2 }}\,du}.
$$
From this we get
$$
\mu _n  = \frac{1}{{n + 1}}\bigg(\frac{{\mu _{n + 2} }}{{x^2 }} - \frac{{\mu _{n + 1} }}{x}\bigg).
$$
Dividing both sides by $x^n$ yields
$$
\frac{{\mu _n }}{{x^n }} = \frac{1}{{n + 1}}\bigg(\frac{{\mu _{n + 2} }}{{x^{n + 2} }} - \frac{{\mu _{n + 1} }}{{x^{n + 1} }}\bigg),
$$
and hence
$$
\frac{{\mu _{n + 2} }}{{x^{n + 2} }} = \frac{{\mu _{n + 1} }}{{x^{n + 1} }} + (n + 1)\frac{{\mu _n }}{{x^n }},
$$
leading to
$$
e^{x + x^2 /2} = \sum\limits_{n = 0}^\infty  {\frac{{\mu _n }}{{n!}}}  = \sum\limits_{n = 0}^\infty  {\frac{{(\mu _n /x^n )x^n }}{{n!}}} .
$$
Finally, letting $a_n = \mu_n / x^n$, we have that
$$
e^{x + x^2 /2} = \sum\limits_{n = 0}^\infty  {\frac{{a_n x^n }}{{n!}}},
$$
with $a_0 = a_1 = 1$, and $a_{n + 2}  = a_{n + 1}  + (n + 1)a_n $, as desired.
A: Let $$S(x)= \sum_{n\ge 0} \frac{a^n}{n!}x^n$$
and note that 
$$S'(x)= \sum_{n\ge 1} \frac{a^n}{(n-1)!}x^{n-1}$$
Replacing in the first equation the value of $a_n$ given by the recursion, $a_{n}=(a_{n+2}-a_{n+1})/(n+1)$ we get
$$S(x) = S_2(x) - S_1(x)$$
where
$$S_1(x) =  \sum_{n\ge 0} \frac{a^{n+1}}{(n+1)!}x^n  =\frac{1}{x} [ S(x)-a_0 ]$$
and
$$S_2(x) =  \sum_{n\ge 0} \frac{a^{n+2}}{(n+1)!}x^n  =\frac{1}{x} [ S'(x)-a_1 ]$$
What gives
$$S(x) \; \left( x+1 \right) = S'(x) + a_0 - a_1 = S'(x)$$
This first order differential  equation is straightforward:
$$\frac{S'(x)}{S(x)} = \log'{S(x)} = x+1 \Rightarrow \log{S(x)} = \frac{1}{2}x^2+x +C$$
which, together with the initial condition $S(0)=1$, gives $S(x)=\exp(x+x^2/2)$.
A: To clarify, let $f(x)=\exp \left( x + \frac{1}{2} x^2 \right)$. Now differentiate $f$ twice, you should end up with a differential equation, something like $f''(x)=(2+2x+x^2)f(x)$ with the initial conditions $f(0)=f'(0)=1$ (see given values of $a_0$ and $a_1=1$). Now you need to solve this ordinary differential equation using the method of power series (most elementry books on ODEs have a chapter on this, you dont need to go too deep into the theory at all, no need for the method of frobenius etc. if you can follow any of the examples in the book you should be able to tackle the question). 
Basically assume $f(x)=\sum_{n\geq 0} a_n x^n$ (noticed i've absorbed the n! term into $a_n$ for simplicity) then substitute into the ODE. By equating the coefficients you should get a recurrence relation for $a_n$. If it matches what you've written in your question, then your done. 
