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### $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12) For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x > 0$. Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and ...
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### Show that $\sqrt{2+\sqrt{2+\sqrt{2…}}}$ converges to 2 [duplicate]

Consider the sequence defined by $a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n > 1$. I know 2 is an upper bound of this sequence (I proved ...
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### Assuming convergence of the following series, find the value of $\sqrt{6+\sqrt{6+\sqrt{6+…}}}$ [duplicate]

Assuming convergence of the following series, find the value of $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ I was advised to proceed with this problem through substitution but that does not seem to help unless ...
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### Find the value of $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }$ [duplicate]

How to prove that $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }=2$
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### Real Analysis Sequence limit [duplicate]

Consider the following sequence: $$\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}} \cdots$$ a) Prove by induction that all terms of the sequence are bounded above by two. b) Show that this ...
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### How to find this limit and prove it rigorously: $\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2…)}}}$? [duplicate]

$\sqrt{2 + \sqrt{2+\sqrt{2+\sqrt2...)}}}$. Pretty classic question, I think - and the limit is equal to 2. But how do I prove this rigorously? An epsilon-delta proof wouldn't work, since I wouldn't ...
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### Construct a set of real numbers whose limit points comprise the set of integers $\mathbb{Z}$

My thought process is the following: Let $S=\{ m + \frac{1}{n}| m \in \mathbb{Z},n \in N \}$. Then I need to show that the limit points of $S$ are indeed the integers and that these are the only ...
### Show that the sequence $\sqrt{2}, \ \sqrt{2+\sqrt{2}}, \ \sqrt{2+\sqrt{2+\sqrt{2+}}}…$ converges and find its limit. [duplicate]
Setting $a_n=\sqrt{2+\sqrt{2+\sqrt{2+}}}\$ I get that $$a_{n+1}=\sqrt{2+a_n} \quad \quad a_1=\sqrt{2}.$$ Clearly all numbers in the sequence are positive and we see that $a_n<a_{n+1} \ \forall \ n$...
Let $a_1=\sqrt2$ and let $a_n=\sqrt{2+a_{n-1}}$ for $n \ge 2$. How do I prove that this sequence converges?