Infinite root 2's

Calculus Level 1

Find the limit of the following sequence:

2 , 2 2 , 2 2 2 , \sqrt2 , \sqrt{2\sqrt2} , \sqrt{2\sqrt{2\sqrt2}} ,\ldots

Image Credit: Flickr Rudolf Vlček .
0 1 2 It diverges

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5 solutions

The sequence of 2 \sqrt{2} goes forever. So the limits of this is:

A = 2 2 2 2 A= \sqrt{2{\sqrt{2{\sqrt{2{\sqrt{2\ldots}}}}}}} A 2 = 2 × 2 2 2 2 \Rightarrow A^2=2\times \sqrt{2{\sqrt{2{\sqrt{2{\sqrt{2\ldots}}}}}}} A 2 = 2 × A A^2= 2 \times A A 2 A = 2 A = 2 \Rightarrow \frac{A^2}{A}=2 \Rightarrow \boxed{A=2}

Well you first have to show the limit exists.

Daniel Liu - 5 years, 9 months ago

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Nice solution but how to show the limit exits , Please help.

Syed Baqir - 5 years, 9 months ago

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the limit exists at 2.

Lipsa Kar - 5 years, 9 months ago

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@Lipsa Kar You'll need to prove that though.

Daniel Liu - 5 years, 9 months ago

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@Daniel Liu correct me if i'm wrong, we need to prove (2x)^1/2 limit exists at 2 or not. {2(2+h)}^1/2= root 2 x root of (2+h)= 2= rhs similarly, lhs=2 and putting value of 2 in it, we get 2. lhs=rhs=f(2). hence, limit exists at 2.

Lipsa Kar - 5 years, 9 months ago

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@Lipsa Kar Sorry, but I don't really understand your explanation. Also, what do you mean by "limit exists at 2"? Do you mean "the limit exists, and it's value is 2"?

Daniel Liu - 5 years, 9 months ago

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@Daniel Liu Can you please give your explanation and method of proving that limit exist.

Syed Baqir - 5 years, 9 months ago

@Daniel Liu what i meant was the limit of that function exists at x=2. where the function is f(x). how do you show the limit exists or not at that point? we consider the left hand limit and the right hand limit. f(x) is there, then to know the left hand limit, which we show as lim (x tends to 2 minus but not exactly 2, just before 2) which we take as f(2-h) where h tends to 0. we then solve that, and after solving everything, i find that, f(2)= f(2-h)=f(2+h) where h tends to 0. i have just applied the old common method. what i meant by limit exists at 2 is that the limit exists at x=2 where the function is f(x). i know my explanation is not that great but i hope i'm clear now. :)

Lipsa Kar - 5 years, 9 months ago

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@Lipsa Kar Ah no, I'm not exactly sure what you are doing and what the function you are talking about is, but I'm fairly sure we are talking about different things.

What I said required proof was this: Given the sequence a 0 = 1 a_0=1 and a n = 2 a n 1 a_n=\sqrt{2a_{n-1}} , prove that lim n a n \lim\limits_{n\to \infty} a_n exists.

Daniel Liu - 5 years, 9 months ago

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@Daniel Liu Can you please go with this limiting from beginning step by step until end , I really love to add that in my Brain...

I want to become Math Champ ! :D

Syed Baqir - 5 years, 9 months ago

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@Syed Baqir See Otto Bretscher's explanation below.

Daniel Liu - 5 years, 9 months ago

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@Daniel Liu Thanks , If you have some other explanation please dont hesitate to share with us.

Cheers, :D

Syed Baqir - 5 years, 9 months ago

@Daniel Liu so sorry, my function wasn't correct! yours is the correct solution! thanks for helping out :)

Lipsa Kar - 5 years, 9 months ago

A^2=2A A=0,2 both. ......... So why is "0" not an answer??

Rams Rahim - 5 years, 9 months ago

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2^x is continuous and as x -> infinity , limit approaches 2 not 0.

Syed Baqir - 5 years, 9 months ago

Just working out the first few terms of the series, the sequence soon exceeds 2 and each successive term gets higher than the previous. The solution offered is asking us to potentially divide infinity by infinity. The series diverges.

Will Benedetti - 5 years, 8 months ago
Otto Bretscher
Aug 26, 2015

The n n th term of the sequence is 2 1 2 + 1 4 + . . . + 1 2 n 2^{\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}} , and the limit is 2 1 = 2 2^1=\boxed{2}

Moderator note:

Nice!

Sir, I am little puzzled because I am wondering how : \sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2.... } } } } } \Longrightarrow { 2 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\quad ...\quad +\quad \frac { 1 }{ { 2 }^{ n } } }

Syed Baqir - 5 years, 9 months ago

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Let's look at the third term of the sequence, 2 2 2 , \sqrt{2\sqrt{2\sqrt{2}}}, for example. Repeatedly using the rule a b = a b \sqrt{ab}=\sqrt{a}\sqrt{b} , we find 2 2 2 = 2 2 4 2 8 = 2 1 2 + 1 4 + 1 8 \sqrt{2\sqrt{2\sqrt{2}}}=\sqrt{2}\sqrt[4]{2}\sqrt[8]{2}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}} . To formally prove the analogous formula 2 1 2 + 1 4 + . . . + 1 2 n 2^{\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}} for the n n th term, use induction. Note that lim n ( 1 2 + 1 4 + . . . + 1 2 n ) = 1 \lim_{n\to\infty}\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}\right)=1 , a geometric series. Now lim n 2 1 2 + 1 4 + . . . + 1 2 n = 2 1 = 2 \lim_{n\to\infty}{2^{\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}}}=2^1=2 since 2 x 2^x is continuous at x = 1 x=1 .

Otto Bretscher - 5 years, 9 months ago

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Thank you.

Syed Baqir - 5 years, 9 months ago

This is the best solution; it takes care of the limit and its existence at the same time.

Daniel Liu - 5 years, 9 months ago
Josh Banister
Aug 27, 2015

Let the sequence be written as a 1 , a 2 , a 3 , a_1, a_2, a_3, \dots . The first term is a 1 = 2 1 2 a_1 = 2^{\frac{1}{2}} and each subsequence term is a n + 1 = 2 a n a_{n+1} = \sqrt{2a_n} . Let a n = 2 b n a_n = 2^{b_n} . Using this, we have 2 b n + 1 = a n + 1 = 2 × a n = 2 × 2 b n = 2 b n + 1 2 b n + 1 = b n + 1 2 \begin{aligned}2^{b_{n+1}} &= a_{n+1} \\ &= \sqrt{2\times a_n} \\ &= \sqrt{2\times 2^{b_n}} \\ &= 2^{\frac{b_n + 1}{2}} \\ \implies b_{n+1} &= \frac{b_n + 1}{2} \end{aligned} .

Now let c n = 1 b n c_n = 1 - b_n . From the above, we get 1 c n + 1 = 1 c n + 1 2 = 1 c n 2 c n + 1 = c n 2 \begin{aligned}1-c_{n+1} &= \frac{1-c_n + 1}{2} \\ &= 1 - \frac{c_n}{2} \\ \implies c_{n+1} &= \frac{c_n}{2} \end{aligned}

Since a 1 = 2 a_1 = \sqrt{2} and hence b 1 = 1 2 b_1 = \frac{1}{2} , then c 1 = 1 2 c_1 = \frac{1}{2} and lim n c n = 0 \lim_{n\to\infty} c_n = 0 . Therefore lim n b n = lim n 1 c n = lim n 1 0 = 1 \lim_{n\to\infty} b_n = \lim_{n\to\infty} 1 - c_n = \lim_{n\to\infty} 1 - 0 = 1 . Finally, this results in lim n a n = lim n 2 b n = 2 lim n b n = 2 1 = 2 \begin{aligned} \lim_{n\to\infty} a_n &= \lim_{n\to\infty} 2^{b_n} \\ &= 2^{\lim_{n\to\infty} b_n} \\ &= 2^1 \\ &= \boxed{2} \end{aligned}

Nice solution but last paragraph is a little vague for me or maybe for some other ppl could you please elaborate more ,

Espiecially , lim Bn = lim (1-Cn) => lim ( 1 - 0 ) = 1

And also , lim An = 2^(Bn)

Thank you very Much

"Upvoted"

Syed Baqir - 5 years, 9 months ago

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Well from before, I defined a n = 2 b n a_n = 2^{b_n} and b n = 1 c n b_n = 1-c_n so I substituted them in.

Josh Banister - 5 years, 9 months ago

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Ok, I got that. Thanks.

Syed Baqir - 5 years, 9 months ago
Hadia Qadir
Aug 30, 2015

The limit pf this sequence converges to 2.

Riccardo Frosini
Aug 25, 2015

By squaring both sides of equation x = 2 2 2 x=\sqrt{2\sqrt{2\sqrt{2\dots}}} and then dividing by two we get: x 2 / 2 = 2 2 2 = x x^2/2=\sqrt{2\sqrt{2\sqrt{2\dots}}}=x . That is x 2 = 2 x x^2=2x and therefore x = 2 x=2 .

x x can be 0 0 too!

Kishore S. Shenoy - 5 years, 9 months ago

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under the root, there is a non-zero value always. and nothing is added or subtracted, things here are just multiplied. we can only get the answer 0 if there was a term 0 in the series. so, we are taking 2 as the answer.

Lipsa Kar - 5 years, 9 months ago

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Can you then explain why 0 came mathematically?

Kishore S. Shenoy - 5 years, 9 months ago

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@Kishore S. Shenoy we are considering the value of that series to be x(say) and then we see that under the root, the same series is being multiplied with 2. mathematically, since x is on both rhs and lhs and it is multiplied, x has a value 0 (x=(2x)^1/2) and 0 satisfies the equation mathematically. but here, in actual, we know that 0 can never be the value of x since all it has is non-zero numbers multiplied with each other. another random example, root of 4 is always 2 but 4 can be written as (-2)(-2). only positive roots are being considered. mathematically true, but not actual

Lipsa Kar - 5 years, 9 months ago

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@Lipsa Kar Oh, that multiplication! Thanks!

Kishore S. Shenoy - 5 years, 9 months ago

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@Kishore S. Shenoy welzzo!! :)

Lipsa Kar - 5 years, 9 months ago

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