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Let $a_1,a_2,..,a_n$ be sequence of real numbers such that $a_{n+1}=a_{n}+\sqrt{1+a_n^2}$ and $a_0=0$.

How to evaluate $\lim_{ n\to \infty }\frac{a_n}{2^{n-1}}$ ?

Martin Sleziak
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    I would rewrite the recurrence in terms of $b_n:=a_n/2^{n-1}$, and see where that leads... I'm afraid it only gives the qualitative conclusions in Yuval's answer. – Jyrki Lahtonen Jun 21 '15 at 17:12
  • @JyrkiLahtonen: didn't you recognize the bisection formula for the cotangent? – Jack D'Aurizio Jun 21 '15 at 17:30
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    Afraid I didn't, @Jack. For some reason I was trying to fit a hyperbolic cosine in there :-( Well done! – Jyrki Lahtonen Jun 21 '15 at 17:32
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    See also: http://math.stackexchange.com/questions/1175041/convergence-of-sequence-given-by-x-1-1-and-x-n1-x-n-sqrtx-n21 and http://math.stackexchange.com/questions/942257/find-the-limit-lim-n-to-infty-k-n-2n-for-k-1-0-and-k-n1-k-n-sqrt1 – Martin Sleziak Jan 31 '17 at 12:35

2 Answers2

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Since: $$ \cot\frac{x}{2}=\cot x+\sqrt{1+\cot^2 x} $$ and $a_0=\cot\frac{\pi}{2}$ we have, by induction: $$ a_n = \cot\frac{\pi}{2^{n+1}} $$ and the wanted limit equals $\displaystyle{\color{red}{\frac{4}{\pi}}}$.

Jack D'Aurizio
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Write $$ a_{n+1} = 2a_n + [\sqrt{1+a_n^2} - a_n], $$ and note that $\sqrt{1+a_n^2} - a_n \approx \frac{1}{2a_n}$. So roughly speaking, $a_{n+1} \approx 2a_n$, so that $a_n/2^{n-1}$ does approach a limit; this needs to be argued more formally, but I leave you the details. There is no particular reason that the limit have any nice expression; calculation shows that it is about 1.27323954473516.

Yuval Filmus
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