<s>[INST] Proof regarding convergence I am supposed to show that the sequence converges and to find the limit.
We've just started going over sequences and induction so I'm very limited in what I know/can use and I was wondering if the general pattern of the proof is okay and if it's correct.. Also it took me quiet a bit of time to prove it this way and I was wondering if there are any shortcuts I can do in the future?
$$ \{a_n\}=\left\{ 
\begin{array}{lcl}
a_1 = 3\\
a_{n+1} = 1+\sqrt{6+a_n}\\
\end{array}
\right.$$
First we show by induction $ \{a_n\} $ is increasing, that is, to prove that the following hold for every $n∈N$
$1+\sqrt{6+a_n}\ge a_n$
$\sqrt{6+a_n}\ge a_n-1$
$6+a_{n}\ge\left(a_{n}+1\right)^{2}$
$6+a_{n}\ge a_{n}^{2}-2a_{n}+1$
$0\ge a_{n}^{2}-3a_{n}-5$
Base case: $n=1$
$1+\sqrt{6+3}\ge 3$
$4\ge 3$
Inductive step: Assume $0\ge a_{n}^{2}-3a_{n}-5$, we show $0\ge a_{n+1}^{2}-3a_{n+1}-5$
$0\ge a_{n}^{2}-3a_{n}-5$
$6+a_{n}\ge a_{n}^{2}-2a_{n}+1$
$\sqrt{6+a_{n}}\ge a_{n}-1$
$0\ge a_{n}-1-\sqrt{6+a_{n}}$
$0\ge1+2\sqrt{6+a_{n}}+6+a_{n}-3-3\sqrt{6+a_{n}}-5$
$0\ge\left(1+\sqrt{6+a_{n}}\right)^{2}-3\left(1+\sqrt{6+a_{n}}\right)-5$
$0\ge a_{n+1}^{2}-3a_{n+1}-5$
Therefore by induction, $\{a_n\}$ is increasing for every $n∈N$.
Now we show by induction $\{a_n\}$ is bounded, that is, $a_n ≤ 4.2$ for every $n∈N$.
Base case: $n=1$, which is trivially true.
Inductive step: Assume $a_n ≤ 4.2$, we show $a_{n+1} ≤ 4.2$.
$a_n ≤ 4.2$
$6+a_{n}\le10.2$
$\sqrt{6+a_{n}}\le\sqrt{10.2}$
$\sqrt{6+a_{n}}\le3.2$
$1+\sqrt{6+a_{n}}\le4.2$
Therefore by induction $\{a_n\}$ is bounded from above.
As $\{a_n\}$ is both bounded and incresing for every $n∈N$, $\{a_n\}$ converges and $\lim _{n\to \infty }\left(a_n\right)=L$ for some $L∈R$
As $\lim _{n\to \infty }\left(a_n\right)=L$, then $\lim _{n\to \infty }\left(a_{n+1}\right)=L$, therefore
$L\ =\ 1+\sqrt{6+L}$
$L\ -1=\ \sqrt{6+L}$
$\left(L\ -1\right)^{2}=\ 6+L$
$L^{2}-2L+1=\ 6+L$
$L^{2}-3L-5=\ 0$
$L = \dfrac{ 3 \pm \sqrt{(-3)^2 - 4*(-5)}}{ 2 }$
$L_1 = 4.19258$
$L_2 = −1.19258$
As $a_n ≥ 0$ for every $n∈N$, $L = 4.19258$
 [/INST] Your proof is honestly very tedious and should be replaced ! You must make use of the recursive property of the sequence directly. Here is how induction works for your problem. Base case: $a_2 = 4 \ge 3 = a_1$ is true. Assume $a_n  \ge a_{n-1}$ for $n \ge 2$, then using the step $a_n \ge a_{n-1} \implies a_{n+1} = 1+\sqrt{6+a_n} \ge 1 +\sqrt{6+a_{n-1}} = a_n$. Hence by induction $a_n \ge a_{n-1}$ for all $n$ and the sequence is monotonically increasing. Next you can show too by induction again that the sequence is bounded above by $5$ and below by $4$. Here again, base case $a_1 = 3 < 5$, assume $a_n < 5$, then $a_{n+1} = 1 +\sqrt{6+a_n} < 1 +\sqrt{6+5} = 1+\sqrt{11} < 1+\sqrt{16} = 1+4 = 5$. Thus $a_n < 5$ for all $n$. Thus the $\{a_n\}$ has a limit $L$ which is the solution of the equation: $L = 1+\sqrt{6+L}$. It can be solve by squaring both sides and note that $4 \le L \le 5$. Can you solve for $L$ ?
</s>