A recursively defined sequence and a limit Fix real numbers $ a_0 $, $ a_1 $ and define, $$ a_{n+1} = a_n +  \Big(\frac{2}{n+1} \Big) a_{n-1} \space \space \forall \space n \ge 1 $$ Show that the sequence $ \Big\{ \dfrac{a_n}{n^2} \Big\}_{n=1}^{ \infty} $ is convergent and find its limit.
This question is from an old Miklos Schweitzer competition (1958, problem 7, as pointed out by @user37238). I don't have any good ideas of approaching this question. Writing the recursion as $ (n+1) (a_{n+1}-a_n)=2a_{n-1} $ gives me the impression that it is a relation between the coefficients of a power series, obtained as a solution to some differential equation, although this line of thought doesn't lead me too far. 
 A: Following Did's comment, let we set:
$$ f(x) = \sum_{n\geq 0} a_n x^n. \tag{1}$$
The trick is that we can turn the recurrence relation into a differential equation for $f(x)$, solve it (or just exhibit an accurate approximation of the solution) and prove the claim. The recursion is:
$$ 2a_{n-1} = (n+1) a_{n+1} - (n+1) a_n \tag{2}$$
but, by setting $a_{-1}=0$:
$$\begin{eqnarray*} \sum_{n\geq 0} a_{n-1}x^n &=& x\cdot f(x) \\ \sum_{n\geq 0} (n+1) a_{n+1} x^n &=& f'(x)\\\sum_{n\geq 0}(n+1)a_n x^n &=& f(x)+x\cdot f'(x)\end{eqnarray*}\tag{3}$$
hence $(2)$ turns into:
$$ (2x+1)\cdot f(x) =   (1-x)\cdot f'(x)\tag{4} $$
that is a separable differential equation, with solutions given by:
$$ f(x) = K \cdot \frac{e^{-2x}}{(1-x)^3} \tag{5}$$
and that is simply wonderful, because $f(x)$ is a meromorphic function with a triple pole at $x=1$.
Since
$$ \frac{1}{(1-x)^3} = \sum_{n\geq 0}\binom{n+2}{2}x^n \tag{6}$$
by stars and bars, and the Laurent series of $f(x)$ in a neighbourhood of its only singularity is given by:
$$ f(x) = \frac{K e^{-2}}{(1-x)^3} + \frac{g(x)}{(1-x)^2}\tag{7}$$
with $g(x)$ being a holomorphic function, we have:

$$ \lim_{n\to +\infty}\frac{a_n}{n^2} = \frac{K}{2e^2} = \color{red}{\frac{a_0}{2e^2}}.\tag{8}$$

