Find limit of $\{a_n\}$ 
The sequence $\{a_n\}$ is determined by $$a_1 = 1, a_{n+1} = \frac{3n-1}{3n} a_n + \frac{1}{n^2}, \quad \forall n\ge 1.$$
Find the limit of $\{a_n\}$ (if it exists).

I guess the limit is $0$ by using MATLAB, but the sequence converges really slowly.
 A: Notice that
$$na_{n+1}-na_n = -\frac{a_n}{3} + \frac{1}{n}
\implies na_{n+1}-(n-1)a_n = \frac{2a_n}{3} + \frac{1}{n}.$$
Hence if the limit of $(a_n)_n$ exists and it is equal to $L\in\mathbb{R}$ then, by Stolz-Cesaro Theorem,
$$L=\lim_{n\to \infty}a_n=\lim_{n\to \infty}\frac{(n-1)a_n}{(n-1)}=
\lim_{n\to \infty}\frac{na_{n+1}-(n-1)a_n}{n-(n-1)}=
\lim_{n\to \infty}\left(\frac{2a_n}{3} + \frac{1}{n}\right)=\frac{2L}{3}$$
and we may conclude that the limit $L$ is zero.
P.S. It is easy to show by induction that $a_n\geq 3/n$ for all $n\geq 2$, which implies that $(a_n)_{n\geq 2}$ is positive and decreasing. Therefore the limit $L$ does exist and $L\in [0,a_2)=[0,5/3)$.
A: If $a_1=1$ and
$$
a_{n+1} = \frac{3n-1}{3n} a_n + \frac{1}{n^2}
$$
Then
$$
\begin{align}
a_n
&=a_1\prod_{k=1}^{n-1}\frac{3k-1}{3k}+\sum_{k=1}^{n-1}\frac1{k^2}\left(\frac{3k+2}{3k+3}\right)\left(\frac{3k+5}{3k+6}\right)\cdots\left(\frac{3n-4}{3n-3}\right)\\
&=a_1\prod_{k=1}^{n-1}\frac{k-\frac13}{k}+\sum_{k=1}^{n-1}\frac1{k^2}\prod_{j=k+1}^{n-1}\frac{j-\frac13}{j}\\
&=a_1\frac{\Gamma\!\left(n-\frac13\right)}{\Gamma\!\left(\frac23\right)}\frac1{\Gamma(n)}+\sum_{k=1}^{n-1}\frac1{k^2}\frac{\Gamma\,\left(n-\frac13\right)}{\Gamma\,\left(k+\frac23\right)}\frac{\Gamma(k+1)}{\Gamma(n)}\\
&=\underbrace{\frac{\Gamma\!\left(n-\frac13\right)}{\Gamma(n)}}_{\sim n^{-1/3}}\left(\vphantom{\sum_{k=1}^{n-1}}\right.\frac{a_1}{\Gamma\!\left(\frac23\right)}+\sum_{k=1}^{n-1}\underbrace{\frac1{k^2}\frac{\Gamma(k+1)}{\Gamma\,\left(k+\frac23\right)}}_{\sim k^{-5/3}}\left.\vphantom{\sum_{k=1}^{n-1}}\right)\\
&=O\!\left(n^{-1/3}\right)
\end{align}
$$
Thus,
$$
\lim_{n\to\infty}a_n=0
$$
