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Calculate $$\lim_{n\to\infty}{\sqrt{5+\sqrt{5+...+\sqrt 5}}}$$

I have a similar example solved in my textbook so I tried to use the same technique though I'm not entirely sure what I'm doing.

$a_n:=\lim_{n\to\infty}{\sqrt{5+\sqrt{5+...+\sqrt 5}}}$

$a_1=\sqrt5$

$a_{n+1}=\sqrt{5+a_n}$

$L=\sqrt{5+L} \implies L=\frac{1+\sqrt{21}}{2}$

This is what I'm confused about, where did $L=\sqrt{5+L}$ come from? And after that I have to prove that $a_n$ is bounded above with $\frac{1+\sqrt{21}}{2}$, right?

So I use induction to prove that:

$a_1=\sqrt5\leq\frac{1+\sqrt{21}}{2}$

Let's assume that for some $n\in \Bbb N$ it's true that $a_n\leq\frac{1+\sqrt{21}}{2}$

Then $a_{n+1}=\sqrt{5+a_n}\leq\sqrt{5+\frac{1+\sqrt{21}}{2}}=\sqrt\frac{11+\sqrt{21}}{2}$

Well, this doesn't seem right. What did I prove here?

After that I proved that $a_n$ is strictly increasing, so I should now be able to find the limit only I don't know how to proceed.

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    If the sequence converges then $\lim a_{n+1}=\lim a_n$ – 2017-01-06
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    $$ L = \sqrt{5+\sqrt{5+\sqrt{5+...+\sqrt 5}}} \Rightarrow L^2 = 5+\sqrt{5+\sqrt{5+...+\sqrt 5}} \Rightarrow \\ L^2-5 = \sqrt{5+\sqrt{5+...+\sqrt 5}} = L \Rightarrow \color{red}{L^2-L-5 = 0} $$ – 2017-01-06
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    So the answer is just $L$. No further justification needed. Also see [here](https://en.m.wikipedia.org/wiki/Nested_radical) under the infinitely nested radicals section. – 2017-01-06
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    @did: the other question does not address the issue of convergence. – 2017-01-06

2 Answers 2

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To the last question, $$ \sqrt{\frac{11+\sqrt{21}}2}=\frac{\sqrt{22+2\sqrt{21}}}2=\frac{\sqrt{1+2\sqrt{21}+21}}2=\frac{1+\sqrt{21}}2 $$ by the binomial theorem.

But you only need any upper bound, you can use $$ a_n\le 4\implies a_{n+1}=\sqrt{5+a_n}\le\sqrt{5+4}=3<4. $$

Obviously from the original expression, the sequence is increasing, which also can be proven using induction. Together both facts imply that the sequence converges, and as the square root is continuous, one has that $$ L=\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{5+a_n}=\sqrt{5+\lim_{n\to\infty}a_n}=\sqrt{5+L}, $$ which then results in a quadratic equation with the additional constraint $L>2$ to select the correct root.

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    Thanks. When in the end I get the quadratic equation, I automatically know that $\frac{1-\sqrt21}{2}$ can't be the correct root since $L$ has to be non-negative, right? So I don't even have to look at the condition $L>2$? – 2017-01-06
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    Yes, of course a negative number will not be larger than 2. – 2017-01-06
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Hints:

$L=\sqrt{5+L}$ comes from the recursive definition of the sequence: $a_{n+1}=\sqrt{5+a_n}$ and the function $f(x)=\sqrt{5+x}$ used in the recursion being continuous.

To prove convergence, all you have to prove is the sequence is monotonic and bounded. Actually, you prove it is increasing. $0$, this ensures convergence.

To prove it is convergent, all you have to prove is $a_n$ lives in an interval on which $f(x)>x$. Studying its variations you can prove the function $f(x)-x$ is decreasing on $[-4,+\infty)$, hence on $[0,+\infty)$., and is $0$ at $\frac{1+\sqrt{21}}2$.