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 .

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    $\begingroup$ It goes through. $\endgroup$ – user284331 Apr 12 '18 at 16:50
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    $\begingroup$ Thanks, @GNUSupporter, for the explanation. I started writing a comment to explain why; but saw your comment, more thorough than mine. Appreciate it! $\endgroup$ – Namaste Apr 12 '18 at 16:54
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    $\begingroup$ Really nice job. It is good. You did every step nicely. $\endgroup$ – fleablood Apr 12 '18 at 17:33
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    $\begingroup$ Now replace 5 by c and see if everything carries through. $\endgroup$ – marty cohen Apr 12 '18 at 19:23
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    $\begingroup$ It's perfect. BTW: If you replace $5$ with $c>0,$ so $a_{n+1}^2=c+a_n$ and $a_1=\sqrt c\; $you can show the sequence is increasing and bounded by showing that $0<a_n<L\implies 0<a_n<a_{n+1}<L,$ where $L>0$ and $L^2-L-c=0$. $\endgroup$ – DanielWainfleet Apr 13 '18 at 1:02

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