<s>[INST] 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!
 [/INST] 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 $S(x)=\exp(x+x^2/2)$.
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