Computing $\lim_{n\to\infty}\sqrt{\frac{2n}{n+1}}$ We are asked to find the limit of the recursively defined sequence, and to assume that the sequence converges. 
$a_1$=0 and $a_{n+1}$= $\sqrt{8+2a_n}$
I then solved for $a_n$ using algebra.
$a_n$=${(a_{n+1})^2 - 8\over 2}$ 
I set the limits of each term equal to eachother.
$\lim \limits_{n \to \infty}$$a_n$ = $\lim \limits_{n \to \infty}$${(a_{n+1})^2 - 8\over 2}$
So, from what I understand
$\lim \limits_{n \to \infty}$$a_n$ = L 
And you can just set
L = ${L^2 - 8\over 2}$ 
which after algebraic simplification equals:
$L^2$ -2L -8 = 0 
(L-4)(L+2) = 0
L = -2, 4
So I do not understand how there are two limits, if these are even correct? And why $(a_{n+1})^2$ can be substituted for $L^2$. I came to these answers after watching a video on a similar problem which is why I'm not really understanding the basic concepts of it. Thanks.
 A: You can prove by induction that this sequence has positive terms, increasing and bounded by 4, and hence converges to a limit L that is positive! Hence 4 is the limit.
We will prove this statement $ 0 \leq a_{n} \leq a_{n+1} \leq 4$ by induction on n. For n=0, the claim is obviously true, since $a_{0}=0$ , $a_{1}= \sqrt{8} \leq 4$. Now we assume that the claim holds for some natural integer n. we show it holds at n+1. Assume that we have for some n, $0 \leq a_{n} \leq a_{n+1} \leq 4$. Multiplying by 2 and then adding 8, $0 \leq 2a_{n}+8 \leq 2 a_{n+1}+8 \leq 16$. Taking squareroots, $0 \leq a_{n+1} \leq a_{n+2} \leq 4$. Hence the claim holds at n+1. By induction, for all n $\in \mathbf{N}$, $0 \leq a_{n} \leq a_{n+1} \leq 4$ which means that $a_{n}$ is increasing positive and bounded (above), and hence converges to L satisfying L = $\sqrt {2L+8}$ and is positive, so $L=4$.
A: It can be done by first solving for $L$. Since the terms are positive, we take $L = 4$. Thus: $0\leq |a_{n+1} - 4| = |\sqrt{8+2a_n}-4| = \left|\dfrac{2(a_n-4)}{\sqrt{8+2a_n}+4}\right| \leq \dfrac{|2(a_n-4)|}{4} = \dfrac{|a_n-4|}{2}< \cdots < \dfrac{|a_0-4|}{2^n}\Rightarrow a_{n+1}-4\to 0 \to a_{n+1} \to 4 = L$, by squeeze theorem.
A: when you wrote $$L = \lim_{n \to \infty} a_{n+1} = \sqrt{8 + 2a_n} \ge 0 $$ it is already a requirement that $L \ge 0.$ that means if you have to choose between/from $-2, 4,$ you will have to take the positive one.
A: When deriving the limit $L$, you removed the square root sign, which opens up the possibility of false solutions (insert $L = -2$ into the original equation $L = \sqrt{8 + 2L}$, and you get $-2 = 2$, which isn't true, but when you eliminate the root by writing $L^2 = 8 + 2L$, you get $4 = 4$, which is true, and you've gained an additional solution).
Another point to raise, which is only tangentially relevant in this specific case is that you arrive at candidate values for $L$ without taking the initial value of the sequence into account in any way. It makes sense that the sequence might converge to a different value if it starts out at a different place. In your case this doesn't happen, but it is a source for multiple solutions in other, similar problems.
You haven't really proven that the limit exists, only that if it exists, then it has to be either $-2$ or $4$. Of course, you can easily see that the sequence never becomes negative, and therefore it is safe to throw away the $L = -2$ solution.
A sequence defined recursively with a nice and simple expression like yours cannot converge to a value that is not a stationary point. What I mean by that is that if we look at the functional expression $f(x) = \sqrt{8 + 2x}$, then the sequence $$0, f(0), f(f(0)), \ldots$$ which is what you have, can only converge to values $L$ such that $L = f(L)$. This gives you $L = \sqrt{8+2L}$, which has the one solution $L = 4$.
As for why we can exchange $(a_{n+1})^2$ with $L^2$, we just assume that we're at a point where $a_n = a_{n+1}$. This will of course never really happen in your sequence, but it is a paraphrasing of the paragraph above.
A: In your proof, you assume that the sequence converges and you call its limit $L$, i.e. $\lim\limits_{n\to \infty}a_n = L$. Clearly then, $\lim\limits_{n\to\infty}(a_{n+1})^2 = \left(\lim\limits_{n\to\infty} a_{n+1}\right)^2 = L^2$. This is due to the fact that $a_n$ and $a_{n+1}$ have the same limits.
What you have proven is, if the limit exists then it is equal to $-2$ or $4$. This is not the same as saying that the limit equals $-2$ and $4$. Because of the square root you can deduce that the limit, if it exists, must be non-negative, hence the only possible answer is $4$.
