whether it converges or not?

Question

Given sequence 〈a_n 〉 n=1 to ∞ defined recursively by a_1=√6,a_(n+1)= √(6+a_n )
Prove that sequence is positive and increasing

Also determine whether sequence is bounded above if so exhibit upper bound if not show that n > 0

Based on conclusion you made above answer whether it converges or diverges

Compute lim as n approaches infinity a_n

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Answer 1

Hmm. Interesting.

The sequence will surely be increasing if it is the case that for all x=a_n for some n, x <(6+x)^(1/2). But if you square both sides of that, you see it's true if and only if x^-x-6<0. And that's true if and only if -23, since above I implicitly assumed x was positive.

So IF the series is increasing, THEN it is surely bounded by 3. Thus, you know what the answer has to be, namely that the series is increasing and bounded by 3.

Well, if x<3, is it also the case that (x+6)^(1/2) <3? Yes indeedy!! It sure is! So you have a quick inductive proof that the sequence is bounded above by 3.

And since every term is <3 and >0, you also know it's increasing.

OK -- that was scrambled, but the key parts of the proof are all there. And it's not hard to put them in nice neat order.

QED (hand-wavingly, at least)

Oh yes -- since it's increasing and bounded, it has a limit. I'd guess that the limit is 3. To prove that it isn't anything lower (it obviously can't be higher), I'd try a proof by contradiction. Assume it's lower. Take a value of a-n that's very close to the limit. Fiddle around and hope to show that a-n+1 is actually over the supposed limit. Since the sequence is increasing, that's enough to prove a contradiction.