If we have for example $a_n=1+\sqrt{a_{n-1}}$ and $\lim_{n \rightarrow \infty} a_n=L$ then can I say that $ L=1+\sqrt{L}$? If it's so, what's the proof?
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Don't you mean $a_n=1+\sqrt{a_{n-1}}$? – Feb 03 '16 at 18:15
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Yes, if $\lim a_n$ exists, then you can use operations like square root, addition, multiplication – SiXUlm Feb 03 '16 at 18:16
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@vrugtehagel corrected – Jelly Belly Feb 03 '16 at 18:16
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2As long as $f$ is continuous this is always true. – Gregory Grant Feb 03 '16 at 18:16
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If the limit of $a_n$ exists and $f$ is continuous, yes. – Eff Feb 03 '16 at 18:16
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Continuity of the square root function is the reason. – Mark Viola Feb 03 '16 at 18:16
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@GregoryGrant Could you provide a demostration if it's possible? thanks! – Jelly Belly Feb 03 '16 at 18:17
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It is enough to suppose $f$ is continuous in $L=\lim a_n$ and $f(a_n)$ is defined for all sufficiently large $n$. – Ángel Valencia Feb 03 '16 at 18:20
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@JellyBelly It's not completely trivial proof to write down, but it's a very standard theorem in any elementary calculus book. I'm sure you can find it. – Gregory Grant Feb 03 '16 at 19:08
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If $f$ is continuous and $a_n$ a converging sequence, then $$f(\lim a_n)=\lim f(a_n)$$ Since $\sqrt{x}$ is continuous, and your sequence bounded, yes, you can say $$\lim a_n=\lim a_{n+1}=\lim (1+\sqrt{a_n})=1+\sqrt{\lim a_n}$$ Thus, $$\lim a_n=\frac 32 + \frac 12\sqrt{5}$$
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To say that $f(\lim a_n)=\lim f(a_n)$ which concept/theorem are you using? – Jelly Belly Feb 03 '16 at 18:23
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3@JellyBelly It is actually a definition of being continuous at $L$. However, you might use another definition (that would be equivalent). See here. – Eff Feb 03 '16 at 18:34
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1@JellyBelly Perhaps it is worth adding caveat that from this we can calculate the limit $L$, assuming the sequence converges. Although in this particular case, for the function $f(x)=1+\sqrt x$, this sequence will converge for any choice of $a_1$. But, for example, if you look at $f(x)=x^2$ and $a_1=2$, then the sequence given by $a_{n+1}=f(a_n)$ clearly does not converge neither to $0$ nor to $1$, which are the only solutions of $f(x)=x$. – Martin Sleziak Feb 03 '16 at 19:37
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Yes, yes you can, since for such recursive definitions limits are fixed points of the function relating a_n to a_{n-1} (as long as it is continuous).
Jsevillamol
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Mainly so we can say that $f(\lim a_n)=\lim f(a_n)=\lim a_{n+1}$. Exchanging the limit and the function can only be done in the case of continuous functions. – Jsevillamol Feb 03 '16 at 18:23