Find limit of $\{a_n\}$

Question

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.

Answer 1

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 $L\in [0,a_2)=[0,5/3)$.

Answer 2

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, $$ \lim_{n\to\infty}a_n=0 $$