If $a_{n+1}=\frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $\lim\limits_{n\to\infty}\left(\frac{4}{3}\right)^n(3-a_n)$?

Question

Question:

If $a_{n+1}=\frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $\lim\limits_{n\to\infty}\left(\frac{4}{3}\right)^n(3-a_n)$ ?

My approach:

I am able to prove separately that the sequence $a_n$ is convergent and $\left(\frac{4}{3}\right)^n$ is divergent.

But I somehow cannot find out how will their multiplication behave..

But as I cant conclude about the convergence of the entire sequence within the required limit, I cannot find the limit.

Please help

Thank you.

Answer 1

EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.

Let's try looking at rations again:

$$ \begin{align} R_n = \frac{(\frac{4}{3})^{n+1}(3-a_{n+1})}{(\frac{4}{3})^{n}(3-a_{n})} &= \frac{4}{3} \times \frac{3-\frac{3+a_n^2}{a_n+1}}{3-a_{n}} = \frac{4}{3} \times \frac{\frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \\ &= \frac{4}{3} \times \frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} = \frac{4}{3} \times \frac{a_n}{a_n+1} \end{align} $$

We know that $a_n < 3$. Since $\frac{x}{x+1}$ is increasing, $\frac{a_n}{a_n+1} < \frac{3}{4}$. Therefore, $R_n < 1$.

Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.

Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.