Is there a general form for $a_n=\sqrt{2+a_{n-1}}$?

Question

Can I find the general term of this sequence $a_n=\sqrt{2+a_{n-1}}$, $a_1=\sqrt2$? I have proved the convergence. And found its limit. But is there any general form for it?

@6005 That's a pretty strong comment to make. You would have to back this up, even with empirical evidence (if proving is too hard, at least support your statements with something). [But the answer below seems to hint that you'd have a hard time doing so.] — Clement C., Dec 11 '16 at 11:06
See also: Proof of an equality involving cosine $\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{2}}}}\ =\ 2\cos (\pi/2^{n+1})$ — Martin Sleziak, Dec 11 '16 at 12:13

score 7 · Answer 1 · answered Dec 11 '16 at 11:04

7

Yes, there is. The formula is $$a_n=2\cos\frac{\pi}{2^{n+1}}.$$ The proof is by induction on $n$, and is left as an exercise for the reader.

answered Dec 11 '16 at 11:04

Jyrki Lahtonen

133,153

wow. How did you get there? – PaulDirac Dec 11 '16 at 11:05
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@DenisNichita I'll risk an answer: by memory. – Git Gud Dec 11 '16 at 11:06
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Experience :-) The recurrence formula (appropriately scaled) is a perfect match with the angle halving formula for cosine. I'm sure the details have been covered many times on our site. Hence CW. – Jyrki Lahtonen Dec 11 '16 at 11:06
Is there any rigorous proof? I don't think you just based this answer by intuition.... – PaulDirac Dec 11 '16 at 11:09
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Well, once you have this guess, you can prove it by induction, as stated. Hard to find, easy to verify. – Clement C. Dec 11 '16 at 11:11
The same trick worked also here. – Jyrki Lahtonen Dec 11 '16 at 18:28

score 3 · Accepted Answer · answered Dec 11 '16 at 11:10

3

Hint: If $a_n=2 \cos(\theta_n)$ with $0 < \theta_n<\pi/2$, then:

$\sqrt{2+a_n}=\sqrt{2+2 \cos(\theta_n)}$

$=2 \sqrt{(1+\cos(\theta_n))/2}$

$=2 \cos((\theta_n)/2)$ using the appropriate half-angle formula.

answered Dec 11 '16 at 11:10

Oscar Lanzi

39,403

Is there a general form for $a_n=\sqrt{2+a_{n-1}}$?

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