# Thread: Limit of sqrt(2*sqrt(2)) sequence

1. ## Limit of sqrt(2*sqrt(2)) sequence

I'm sure you people know which sequence I'm talking about. My teacher said it was popular. I'm just wondering if someone knows how to find the answer for this without just plugging in values with the calculator which is all I find myself being able to do with this problem.

Any input would be appreciated!

P.S. The limit (=answer) is 2.

2. Originally Posted by s3a
I'm sure you people know which sequence I'm talking about. My teacher said it was popular. I'm just wondering if someone knows how to find the answer for this without just plugging in values with the calculator which is all I find myself being able to do with this problem.

Any input would be appreciated!

P.S. The limit (=answer) is 2.
Let $a_1=\sqrt2,\;a_2=\sqrt{2\sqrt2},\;\cdots,\; a_n=\underbrace{\sqrt{2\sqrt{2...\sqrt2}}}_{n}$

We then get $a_{n+1}=\underbrace{\sqrt{2\sqrt{2...\sqrt2}}}_{n+ 1} = \sqrt{2a_n}$ or in other words $a_{n+1}^2=2a_n$

So if $L$ is our limit then it satisfies $L^2=2L \implies L=0\text{ or }2$.

Show $\{a_n\}_{n=1}^\infty$ is an increasing sequence and that'll tell us $L\neq 0\implies L=2$.

3. Alternatively,

$\sqrt{2\sqrt{2\sqrt{2...}}}=2^{\frac{1}{2}}2^{\fra c{1}{4}}2^{\frac{1}{8}}....$

$=2^{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}....)}$

In brackets is a geometric series, first term=0.5, ratio=0.5.

$Limit=2^{S_{\infty}}=2^{(\frac{0.5}{1-0.5})}=2^{\frac{0.5}{0.5}}=2^1=2$

4. chiph588@'s method assume that sequence has a limit. Of course that can be proven by showing that the sequence is increasing and has 2 as an upper bound.

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# sqrt(2sqrt(2sqrt(2)))

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