Question

Consider the following sequence:
\[\sqrt{2},\sqrt{2\sqrt{2}}\,\sqrt{2\sqrt{2\sqrt{2}}}\]
Prove that it converges, and find it's limit.
EDIT: Sorry, wrong question

Answer 1

Find \(\Large\displaystyle\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}\)

Answer 2

well, did you find the general equation for a(n) ?

Answer 3

@sourwing my guess is: a(n)=sqrt(n sqrt(a(n-1))) with a(1)=sqrt(n)

Answer 4

Bear in mind that \(\sqrt n=n^{0.5}\)

Answer 5

well, that's the recursive formula. I was talking about explicit formula

Answer 6

how about let's just start it.

a(1) = 2^(1/2)
a(2) = 2^(1/2) * 2^(1/4)
a(3) = 2^(1/2) * 2^(1/4) * 2^(1/8)

see the pattern?

Answer 7

yes I see the pattern

Answer 8

so what do you think a(n) is?

Answer 9

I don't know how to state it as a explicit formula, as I don't have much experience with sequences. However, I can clearly see the pattern and state the recursive formula as I did above.

Answer 10

@sourwing don't let him mess with \(\Large\displaystyle\Pi\)

Answer 11

that's fine. These type of problems aren't easy to begin with anyways.

notice that:
a(3) = 2^(1/2 + 1/4 + 1/8) using exponent rule

then a(n) = 2^(1/2 + 1/4 + 1/8 + ... + 1/2^n) yes?

@kc_kennylau i'm not :D

Answer 12

@kc_kennylau as a matter of fact, i'm a noob when it comes to pi product XD

Answer 13

yes, i see it now

Answer 14

good, now can you write it in terms of summation notation?

Answer 15

we haven't talked about summation in class, but would be my guess:
\[\sum_{i=m}^{n} 2^{1/2^{m}}+2^{1/2^{m+1}}....2^{1/2^{n}}\]

Answer 16

I meant to say: \[\sum_{i=m}^{n}2^{1/2^i}\]

Answer 17

you can't do that because 2^(a + b) isn't 2^a + 2^b.

the summation is on the exponent only.
so \[a(n) = 2^{\sum_{i=1}^{n} (1/2)^i}\]

yes?

Answer 18

that's a letter i in case you can't read it

Answer 19

ya that makes sense, I see what I did wrong

Answer 20

Now, technically this is only a propose for the formula of a(n). Even though it's obviously true, we still need to prove that is true for all natural n.

Answer 21

any idea which method of proving we are going to use?

Answer 22

induction

Answer 23

bingo!

suppose it's true for a(k) for some k, then:


yes?