If we sum the given equation for $n=3,4,5, \ldots, N$, we obtain $$\sum_{n=3}^{N} A_{n}=\sum_{n=3}^{N} \frac{A_{n-1}+A_{n-2}+A_{n-3}}{3}+\frac{1}{n^{4}-n^{2}}$$ This reduces dramatically to $$\begin{equation*} A_{N}+\frac{2 A_{N-1}}{3}+\frac{A_{N-2}}{3}=A_{2}+\frac{2 A_{1}}{3}+\frac{A_{0}}{3}+\sum_{n=3}^{N} \frac{1}{n^{4}-n^{2}} \tag{*} \end{equation*}$$ Let $\lim _{N \rightarrow \infty} A_{N}=L$. Under this limit, the left hand side of $\left(^{*}\right)$ is simply $2 L$. We compute the sum on the right with the help of partial fractions $$\begin{aligned} \lim _{N \rightarrow \infty} \sum_{n=3}^{N} \frac{1}{n^{4}-n^{2}} & =\sum_{n=3}^{\infty} \frac{1}{n^{2}-1}-\frac{1}{n^{2}} \\ & =\left(\sum_{n=3}^{\infty} \frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right)+\frac{1}{1^{2}}+\frac{1}{2^{2}}-\sum_{n=1}^{\infty} \frac{1}{n^{2}} \\ & =\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)+\frac{5}{4}-\frac{\pi^{2}}{6} \\ & =\frac{5}{3}-\frac{\pi^{2}}{6} \end{aligned}$$ With this we easily find $L=\frac{1}{2} \cdot\left(2+\frac{2}{3} \cdot 1+\frac{1}{3} \cdot 0+\frac{5}{3}-\frac{\pi^{2}}{6}\right)=\frac{13}{6}-\frac{\pi^{2}}{12}$, and we are done.