I came across the recurrence relation
$$a_{n + 1} = \dfrac{2n - 1}{2n + 1} a_{n - 1} - 2 \dfrac{n^2 + 1}{2n + 1} a_{n}$$
for $n \geq 2$ and $a_1 \in \mathbb{R}$ is arbitrary and $a_2 = - \dfrac{5}{3} a_1$.
After computing a few terms of the sequence, it seems that the sequence is unbounded and hence cannot be convergent. Is there some analytic method to prove my claim?
Note: The sequence is not constant in sign. It may take both positive and negative values. Also, due to this, it is not monotone.
First, it's clear that if $a_1=0$ then the sequence has limit $0$. Now assume $a_1\ne0$.
You can easily check that $a_3,a_4$ have opposite signs and that $$|a_4|\ge4|a_1|\ .$$ We shall show by induction that if $n\ge4$ then $a_{n-1},a_n$ have opposite signs and $|a_n|\ge n|a_1|$, so the sequence diverges.
So, if $a_{n-1}>0$ and $a_n<0$ then the recurrence gives $a_{n+1}$ as a sum of two positive terms, so it is positive. Similarly, if $a_{n-1}<0$ and $a_n>0$, then $a_{n+1}$ is negative. This proves the claim about opposite signs.
Now let $n\ge4$. Then it is easy to check that $$\frac{2n^2+2}{2n+1}\ge2\ .$$ We also know that $$\frac{2n-1}{2n+1}a_{n-1}\quad\hbox{and}\quad -\frac{2n^2+2}{2n+1}a_n$$ have the same sign, so $$\eqalign{|a_{n+1}| &=\frac{2n-1}{2n+1}|a_{n-1}|+\frac{2n^2+2}{2n+1}|a_n|\cr &\ge \frac{2n^2+2}{2n+1}|a_n|\cr &\ge2n|a_1|\cr &\ge(n+1)|a_1|\ .\cr}$$ This completes the proof.