Show the sequence $\bigg\{{\sqrt{5}}~,{\sqrt{5+{\sqrt5}}}~,\sqrt{5+\sqrt{5+\sqrt{5}}}~,...\bigg\}$ converges and find its limit.
Attempt :
Let $a_{1}=\sqrt{5}$ and $a_{n+1}={\sqrt{5+a_{n}}}$ for $n=2,3,...$
Now we apply the induction to prove the sequence$(a_{n})$ is increasing , bounded above by $3.$
$(a_{n})$ is increasing $:$
Note first that $a_{n}\ge0$ for each $n\in{\bf N}$.
As $n=1$, one has $a_{1}^{2}=5<5+\sqrt 5=a_{2}^{2}\Longrightarrow |a_{1}|<|a_{2}|\Longrightarrow a_{1}<a_{2}$ .
Now assume $a_{n-1}<a_{n}$ for some $n\in\bf N$ .
Then $a_{n}^{2}=5+a_{n-1}<5+a_{n}=a_{n+1}^{2}\Longrightarrow |a_{n}|<|a_{n+1}|\Longrightarrow a_{n}<a_{n+1}$ .
$(a_{n})$ is bounded above by 3 $:$
As the same manner , we have $a_{1}=\sqrt 5<\sqrt 9=3$ for $n=1$.
Suppose the process holds for some integer $n>0,$ that is , $a_{n}<3$ for some $n\in\bf N$ .
Whence , $a_{n+1}=\sqrt{5+a_{n}}<\sqrt{5+3}<\sqrt9=3 .$
Therefore, we see on account of the monotonic sequence Theorem that $(a_{n})$ is convergent .
Limit of $(a_{n}):$
Take $\displaystyle\lim_{n\rightarrow\infty}a_{n}=L$ for some $L\ge0$ since $a_{n}$ is non-negative for all $n\in\bf N$.
Thus , $\sqrt{5+L}=\sqrt{5+\displaystyle\lim_{n\rightarrow\infty}a_{n-1}}\color{red}=\displaystyle\lim_{n\rightarrow\infty}\sqrt{5+a_{n-1}}=\displaystyle\lim_{n\rightarrow\infty}a_{n}=L$ , keep in mind that the red equality holds by $\sqrt{x}$ is continuous .
Then , one has $L^{2}-L-5=0\Longrightarrow L=\displaystyle\frac{1+\sqrt{21}}{2}$ since $L$ is non-negative .
Can anyone check my proof for validity if you have the time , otherwise ignore this, that is okay . Any comment or valuable suggestion I will be grateful .
