I am having trouble proving the convergence and finding the limit of the sequence $\left(a_{n}\right)$ given by $a_{1}=\sqrt{3}$ and $a_{n}=\sqrt{3+2 a_{n-1}}$. I need to show that the sequence is convergent and determine its limit.
Here is my attempt at solving the problem:
Firstly, I noticed that the sequence is increasing because $a_{n}=\sqrt{3+2 a_{n-1}}>a_{n-1}$ for all $n>1$. To show that 3 is an upper bound, I started by assuming that $a_{n} \leq 3$ for all $n$. Then, I squared both sides of the recurrence relation and simplified to get:
$$a_{n}^{2}-2 a_{n-1}=3$$
Since $a_{n-1} \geq a_{1}=\sqrt{3}$, we have:
$$a_{n}^{2}-2 \sqrt{3} \leq 3$$
Solving for $a_{n}$, we get:
$$a_{n} \leq \sqrt{6}$$
Therefore, $3 \leq a_{n} \leq \sqrt{6}$ for all $n$, and since the sequence is increasing, it is bounded and therefore convergent.
However, I am stuck on how to determine the limit of the sequence. I would appreciate any help or hints on how to proceed with finding the limit.
Your reasoning is incorrect. How do you know that $a_n$ increases? Why $\sqrt{3+2a_{n-1}}>a_n$? And, you already assumed that $a_n\le 3$ for any $n$, which is what you wanted to show. One may solve as follows.
(1) I claim that for any $n$, $1\le a_n\le 3$. This holds for $n = 1$, and if this holds for $n = k$, then $5\le 3+2a_n\le 9$, and $\sqrt{5}\le\sqrt{3+2a_n}\le 3$, so that $\sqrt{5}\le a_{n+1}\le 3$ and the same bound holds for $n = k+1$.
(2) $a_n$ is increasing. Solving $\sqrt{3+2a_n}\ge a_n$ gives $3+2a_n\ge a_n^2$, $(a_n-1)^2\le 4$, which is correct since $1\le a_n\le 3$.
Hence the sequence is bounded and monotone, so converges. Then the sequence $a_{n+1} = \sqrt{3+2a_n}$ must converge to the same limit $a$, and by the limit laws we have $a = \sqrt{3+2a}$. Solving this gives $a = -1$ or $3$, and $a=-1$ is impossible since the sequence increases and $a_1 = \sqrt{3}$. Thus it must converge to $3$.