$a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ implies $a_n/n^2$ converges? Let $a_0=\pi$, $a_1=\pi^2$, $a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}$ How can we prove that $a_n/n^2$ converges?
It is easy to see that
$$a_n-a_1=\sum_{k=1}^{n-1}\frac{2a_{k-1}}{k+1}\geq 2a_0\sum_{k=1}^{n-1}\frac{1}{k+1}\to\infty.$$
Then how to do?

Let $(b_n)=(\frac {a_n} {n^2})$, we have that
\begin{align}
b_{n+1}-b_n&=\frac {n^2a_{n+1}-(n+1)^2a_n}{n^2(n+1)^2}\\
&=\frac {n^2(a_{n+1}-a_n)-(2n+1)a_n}{n^2(n+1)^2}\\
&=\frac {\frac {2n^2}{n+1}a_{n-1}-(2n+1)a_n}{n^2(n+1)^2}\\
&=\frac {2n^2a_{n-1}-(2n+1)(n+1)a_n}{n^2(n+1)^3}\\
&=\frac {2n^2(a_{n-1}-a_n)-(3n+1)a_n}{n^2(n+1)^3}\\
&\le0
\end{align}
Since $(b_n)\geq0$ and $(b_n)$ is decreasing, $(b_n)$ is convergent.
