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For the two following sequences I want to find their limits:

(1) The sequence $2$, $2\sqrt{2}$,$2\sqrt{2\sqrt{2}}$,...

(2) $a_{n+1}$ = $\sqrt{1+a_n}$, $a_1 = 1$

For both sequences I want to show that they are bounded and monotone increasing.

My ideas for (1): (1) can be written as $a_{n+1} = 2\sqrt{a_n}$. Also I assume that for some n $a_n < 4$ then $a_{n+1} = 2 \sqrt{a_n} < 2 \sqrt{4} < 4$, hence I have 4 as an upper bound (and limit?). How do I show the sequence is monotonic increasing?

For (2) - am I wrong our is the sequence not upper bounded, and hence also not convergent?

3 Answers3

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For two:

First you can prove by induction that $a_n > 1$

Looking for limit point, they have to verify : $L^2 - L -1 = 0$ which gives you only one value >1 (noted g here)

Now, let's prove that $a_n \leq g$ by induction:

$a_1 = 1 < g$ ; if $a_n \leq g$ :

$a_{n+1} = \sqrt{1 + a_n} \leq \sqrt{1 + g} = g $

Now there is a lemma that says : If $a_n$ is bounded and has only one limit point (here g), then it converges towards this limit point. You can prove that by supposing it isn't true: then there is an infinite number of term of this sequence that are distant of a certain value of g, and from this extracted sequence you can find a different limit point (bolzano-weierstrauss theorem regarding bounded sequence) since you can reextract a converging sequence whose limit is by definition different from g. This is absurd because there is only one limit point, hence your sequence converges to its only limit point.

So $(a_n)$ converges

mvggz
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For your first problem, note that we are looking at $2^{e_n}$, where the exponents $e_n$ are $1,1+\frac{1}{2}, 1+\frac{1}{2}+\frac{1}{4}$, and so on.

It is a familiar fact that $\lim_{n\to\infty} e_n=2$.

So we can find an explicit expression for the limit. If we merely want to show boundedness, note that the exponents are always $\lt 2$.

André Nicolas
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  • what do you mean by "exponents"..? –  Oct 26 '14 at 23:37
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    Our sequence can be thought of as $2^{e_1}$, $2^{e^2}$, and so on. The $e_k$ are the exponents. The first number is $2^1$. The second number is $2^{1+\frac{1}{2}}$. The third number is $2^{1+\frac{1}{2}+\frac{1}{4}}$. And so on. – André Nicolas Oct 26 '14 at 23:41
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For (1), show that $0 < x < 4 \Rightarrow 2\sqrt{x} > x$.

For (2), you are wrong. The sequence is bounded above. I would suggest graphing the functions $f(x) = x$ and $g(x) = \sqrt{1+x}$ together to get an idea of what's going on.

If you know $a_n$, you can get $a_{n+1}$ graphically in the following way. Start from the point $(a_n,a_n)$ on the line $y = x$. Move vertically along the line $x = a_n$ until you hit the graph of $g$. That point is $(a_n,a_{n+1})$. Then move horizontally from there along the line $y = a_{n+1}$ until you hit the line $y = x$. That point is $(a_{n+1},a_{n+1})$.

By repeating this process, you can follow the sequence $a_n$ graphically, and guess what its limit is. This can be helpful before trying to prove the facts you want.

user187373
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