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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.

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  • $\begingroup$ The precise reference is apparently Miklós Schweitzer Competition 1958 Problem 7 (see here). $\endgroup$ – user37238 Nov 10 '15 at 14:30
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    $\begingroup$ Hint: Identify the series $$A(t)=\sum_{n=0}^\infty a_nt^n$$ and check that, when $t\to1$, $$(1-t)^3A(t)\to2\ell,$$ for some explicit $\ell$ depending on $(a_0,a_1)$, to deduce finally that $$\frac{a_n}{n^2}\to\ell.$$ The final answer might be $$\ell=2a_0+\frac18\left(1-\frac5{e^2}\right)(a_1-a_0).$$ $\endgroup$ – Did Nov 10 '15 at 18:11
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    $\begingroup$ @Did, why not add your comment as an answer? $\endgroup$ – lhf Nov 10 '15 at 18:25
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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:

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

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