I am trying to establish the boundedness of the following real-valued sequence: Given $\alpha \in \left(0,1 \right)$ and $\beta \in \mathbb{R}$, define $$ \left\{ \begin{array}{l} x_{1} = \alpha \beta , \, x_{2} = 2\alpha^{2} \beta, \\ \displaystyle x_{n+1} = \frac{2n-1}{n} \cdot \alpha \cdot x_{n} - \frac{n-2}{n-1} \cdot x_{n-1} \quad \text{for } n\geq 2. \end{array} \right. $$
I tried to plot the first 1000 terms by using Maple with some concrete numbers $\alpha$ and $\beta$, and my guess is that the sequence above is bounded. In addition, I tried to use the triangle inequality to prove the boundedness, but it was stuck.
Now my attempt is to guess the bound and then use induction; however, it seems to be hard in this situation.
Any help would be appreciated. Regards.
I would expect the behaviour of the sequence to depend on the choice of $\alpha$ and $\beta$.
As @David remarked, start with $\beta = 1/\alpha$. Let us set $\alpha > \frac 43$. Then $x_1 = 1$ and $x_2 = 2\alpha > 1 = x_1$. Furthermore, we can make the following estimate, using $\frac{2n-1}{n} \geq \frac 32$ and $\frac{n-2}{n-1} \leq 1$, valid for $n\geq 2$. $$ x_{n+1} \geq \frac 32 \alpha x_n - x_{n-1}\,. $$ Then $$ x_{n+1} - x_n \geq \left(\frac 32 \alpha - 1\right) x_n - x_{n-1} \geq x_n - x_{n-1}\,,$$ because $\frac 32 \alpha - 1 \geq 1$ and by induction it follows that $x_n$ is monotone increasing. We can then estimate $$ x_{n+1} \geq \frac 32 \alpha x_n - x_n = \left(\frac 32 \alpha - 1\right) x_n \,,$$ which shows that the sequenc $(x_n)$ grows faster than a geometric sequence and hence is unbounded.
I would expect the behaviour to depend on the choice of $\alpha$ and $\beta$. The following is guessing: for large $n$, I would expect the sequence to behave like $$ x_{n+1} = 2\alpha x_n - x_{n-1}\,,$$ and for this sequence you can find an explicit formula. This could give you intuition for proving boundedness/unboundedness of your sequence.