How do I prove using induction that an infinitely nested radical, like sqrt(1+sqrt(1+sqrt(1+... is increasing. I have seen there are many examples on here like this but haven't seen one that proves the sequence is increasing. Could someone give me some tips or link to one which shows the proof that it is increasing? Thanks.
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Start with $a_1=1,a_{n+1}=\sqrt{a_n+1}$, consider various comparisons between $a_n$ and $a_{n+1}$ like the difference test, the ratio test, etc... – abiessu Nov 27 '13 at 02:14
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Define $a_{n+1}=\sqrt{a_n+1}$ with $a_1=1$ as in @abiessu's comment. Then you have to prove that $a_n<a_{n+1}$ for all $n$. For $n=1$ is trivial. Suppose that the result is valid for $n$. That is $$a_n< a_{n+1}.$$ And consider the case $n+1$, so we have to prove that $a_{n+1}< a_{n+2}$ but
$$a_{n+1}< a_{n+2}$$
$$\Leftrightarrow a_{n+1}^2< a_{n+2}^2$$ wich is true since $$a_n+1< a_{n+1}+1$$ by induction hypothesis.
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Let $f(x) = \sqrt{x+1}$, which is an increasing function of $x$ (for $x > -1$).
The composition of increasing functions is increasing.
Let $a_0 = 1$ and $a_{n+1} = f(a_n)$. Thus $a_n = f^{(n)}(a_0)$ (where $f^{(n)}$ is $f$ composed with itself $n$ times), but
$a_{n+1} = f^{(n+1)}(a_0) = f^{(n)}(a_1)$, so since $a_1 = \sqrt{2} > a_0$ we have
$a_{n+1} > a_n$.
On the other hand, if you started from some other $a_0$ so that $a_1 < a_0$, the sequence would be decreasing, not increasing. The boundary between increasing and decreasing is the fixed point, $(\sqrt{5}+1)/2$.
Robert Israel
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