Does the sequence converge?

Calculus Level 3

Define a sequence $\displaystyle{\{a_n\}}_{n=1}^{\infty}$ by $\displaystyle{a_1 = \sqrt{2}}$ $\displaystyle{a_{n+1}} = (\sqrt{2})^{a_n},n \in N$

Find the limit of this sequence.

The answer is 2.

1 solution

Abhishek Singh
Nov 1, 2014

Since $\displaystyle{x_2=(\sqrt{2})^{\sqrt{2}}} >\sqrt{2} = x_1$ and for any integer $\displaystyle{n>1}$ , $x_n > x_{n+1} \Rightarrow (\sqrt{2})^{x_n} > (\sqrt{2})^{x_n+1}$ It follows by induction that the above sequence is an increasing sequence.Also by induction it is easy to see that the sequence is bounded above by 2.Therefore this sequence has a limit, say l .Now the defining relation of the sequence gives $l= (\sqrt{2})^l = 2^{^l/_2} = (1+1)^{^l/_2} \ge 1+ \dfrac{l}{2}$ therefore $\dfrac{l}{2} \ge 1 \Rightarrow l \ge 2$ Hence $\boxed{l = 2}$

${ a }_{ n }\quad =\quad { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }^{ { \sqrt { 2 } }\dots \dots \dots \dots } }\quad N\quad times }\quad \quad \\ \\ L\quad =\quad \lim _{ n\rightarrow \infty }{ { \quad a }_{ n }\quad } \\ \quad \quad \\ \quad L\quad =\quad { (\sqrt { 2 } ) }^{ L }\\ \\ \cfrac { \ln { (L) } }{ L } \quad =\quad \cfrac { \ln { (2) } }{ 2 } \\ \\ \Rightarrow \quad L\quad =\quad 2$ .

But if we draw graph of $\\ f\left( x \right) \quad =\quad \cfrac { \ln { (x) } }{ x } \quad$ .

and then draw a line of $y\quad =\quad \cfrac { \ln { (2) } }{ 2 }$ .

Then this line will intersect at two points. ${ x }_{ 1 }\quad \& \quad { x }_{ 2 }$ .

such that

${ x }_{ 1 }\quad =\quad 2\quad <\quad e\\ \\ { x }_{ 2 }\quad =\quad 4$ .

So There is Two possible answer of this question..!!

Deepanshu Gupta - 6 years, 7 months ago

@DEEPANSHU GUPTA I disagree with your claim that there are 2 possible answers to the question. There is at most 1 limit of a given sequence, and hence at most 1 answer to the question.

What you have done is calculated the "possible" answers for the value that it converges to. This does not mean that that sequence must converge to all of the possible answers. The place where your reasoning broke down is that while the answer must satisfy the equation, not all solutions to the equation is the answer to the problem.

For example, consider the following question: Define a sequence by $\{ a_n \} _{n=1 } ^ \infty$ by $a_1 = 1$ and $a_{n+1} = a_n^2$ . Find the limit of the sequence.

By your argument above, set $L$ as the limit, then we need to solve for $L = L^2$ , which is true for $L = 0, 1$ . Does this mean that the limit of $1, 1, 1 ,1, 1$ is 0?

Note: I have deleted the rest of the comment chain, which is not relevant.

Calvin Lin Staff - 6 years, 7 months ago

Agree with you sir!! But If you asked me to solve your question Then My approach is ${ a }_{ 1 }=\quad 1\\ { a }_{ 2 }=\quad 1\\ { a }_{ 3 }=\quad 1\\ .\\ .\\ .\\ \\ { a }_{ n }\quad =\quad 1\\ \\ L\quad =\quad \lim _{ n\rightarrow \infty }{ { \quad a }_{ n } } \quad =\quad 1\quad$ .

I didn't say that this relation is true for every sequence that is

$\\ \lim _{ n\rightarrow \infty }{ { \quad a }_{ n } } \quad =\quad \lim _{ n\rightarrow \infty }{ { \quad a }_{ n+1 } }$ .

But I say That in the Abhishek's question it should be true if 'n' tends to infinite That it is true for his question Since If i decrease one $\sqrt(2)$ in the sequence when there are infinitely many terms then there is no effect in the limiting Value of this sequence. From This conclusion I write the equation :

$L\quad =\quad { (\sqrt { 2 } })^{ L }\quad$ .

Because In this case (In this Particular Abhishek's Question):

$\lim _{ n\rightarrow \infty }{ { \quad a }_{ n } } \quad =\quad \lim _{ n\rightarrow \infty }{ { \quad a }_{ n+1 } }$ .

This relation is correct !!

And I agree with your claim That Limit to an sequence Should be unique ! But still I didn't Find The mistake in my solution! Correct me where I'am going Wrong ! @Calvin Lin sir!

Deepanshu Gupta - 6 years, 7 months ago

@Deepanshu Gupta – The limit satisfies the equation. <--- This is correct.

All values that satisfies the equation are the limit. <--- This is incorrect.

If x is a limit, then it would satisfy the equation. <-- Correct.

If x satisfies the equation, then x is a limit. <-- Wrong.

Affirming-the-consequent fallacy

Kenny Lau - 5 years, 8 months ago

Does the sequence converge?

The answer is 2.

1 solution

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