How to calculate the limit of $\frac{a_n}{n^2}$ for the sequence $a_{n+1}=a_n+\frac{2 a_{n-1}}{n+1}$?

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 $$ \frac{9\pi-5\pi^2}{8e^2}+\frac{\pi^2-\pi}{8}\approx \color{red}{0.4845}.\tag{4}$$