<s>[INST] How to calculate the limit of $\frac{a_n}{n^2}$ for the sequence $a_{n+1}=a_n+\frac{2 a_{n-1}}{n+1}$? Assume $\{a_n\}$is a sequence which  satisfies the following recursion formula:
$$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}~(n\geqslant 1).$$
and $a_0=\pi,a_1=\pi^2 $.
How to compute $\displaystyle\lim_{n\to \infty}\frac{a_n}{n^2}$?
 [/INST] Let we assume that
$$ f(x) = \pi+\pi^2 x +a_2 x^2+\ldots = \sum_{n\geq 0}a_n x^n \tag{1}$$
The recursion turns into the differential equation
$$\frac{f(x)-\pi-\pi^2 x}{x}=f(x)-\pi+\frac{2}{x}\int_{0}^{x}t\,f(t)\,dt \tag{2} $$
and assuming $G'(x)=x\,f(x)$, that leads to 
$$ f(x) = \frac{\pi(9-5\pi)}{4}\cdot\frac{e^{-2x}}{(1-x)^3}+\frac{\pi(\pi-1)}{4}\cdot\frac{2x^2-6x+5}{(1-x)^3} \tag{3}$$
so $f$ is a meromorphic function with a triple pole at $x=1$. By computing the MacLaurin expansion of $f(x)(1-x)^3$ at $x=1$, it follows that the wanted limit equals
$$ \frac{9\pi-5\pi^2}{8e^2}+\frac{\pi^2-\pi}{8}\approx \color{red}{0.4845}.\tag{4}$$
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