# infinite sequence

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• Dec 6th 2011, 11:01 PM
jaqueh
infinite sequence
What does the series converge to where
[a_1]=√12
[a_n+1]=√(12+a_n) ?

I tried:
L = √(12+√(12+√(12+...)))
then L^2 = 12+√(12+√(12+...))
then L^∞ = 12^∞ + 12^∞-1...+12
thus L = ∞√(12^∞ + 12^∞-1...+12) but i dont think that this is right.

Then someone on physics forum told me that a_n can be written as √(12+a_n+1) so a_n+1 converges because it is in the series itself, but what is the limit?
• Dec 6th 2011, 11:49 PM
CaptainBlack
Re: infinite series
Quote:

Originally Posted by jaqueh
What does the series converge to where
[a_1]=√12
[a_n+1]=√(12+a_n) ?

I tried:
L = √(12+√(12+√(12+...)))
then L^2 = 12+√(12+√(12+...))
then L^∞ = 12^∞ + 12^∞-1...+12
thus L = ∞√(12^∞ + 12^∞-1...+12) but i dont think that this is right.

Then someone on physics forum told me that a_n can be written as √(12+a_n+1) so a_n+1 converges because it is in the series itself, but what is the limit?

If this converges the limit should satisfy:

$L=\sqrt{12+L}$

squaring:

$L^2=12+L$

rearranging:

$L^2-L-12=0$

This has exactly one positive root and it is the limit you seek.

CB
The text passage contains a mathematical question about the convergence of an infinite sequence, and an answer is provided in the form of a solution to the problem.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# infinite sequence

Printable View

• Dec 6th 2011, 11:01 PM
jaqueh
infinite sequence
What does the series converge to where
[a_1]=√12
[a_n+1]=√(12+a_n) ?

I tried:
L = √(12+√(12+√(12+...)))
then L^2 = 12+√(12+√(12+...))
then L^∞ = 12^∞ + 12^∞-1...+12
thus L = ∞√(12^∞ + 12^∞-1...+12) but i dont think that this is right.

Then someone on physics forum told me that a_n can be written as √(12+a_n+1) so a_n+1 converges because it is in the series itself, but what is the limit?
• Dec 6th 2011, 11:49 PM
CaptainBlack
Re: infinite series
Quote:

Originally Posted by jaqueh
What does the series converge to where
[a_1]=√12
[a_n+1]=√(12+a_n) ?

I tried:
L = √(12+√(12+√(12+...)))
then L^2 = 12+√(12+√(12+...))
then L^∞ = 12^∞ + 12^∞-1...+12
thus L = ∞√(12^∞ + 12^∞-1...+12) but i dont think that this is right.

Then someone on physics forum told me that a_n can be written as √(12+a_n+1) so a_n+1 converges because it is in the series itself, but what is the limit?

If this converges the limit should satisfy:

$L=\sqrt{12+L}$

squaring:

$L^2=12+L$

rearranging:

$L^2-L-12=0$

This has exactly one positive root and it is the limit you seek.

CB
user
The text passage contains a mathematical question about the convergence of an infinite sequence, and an answer is provided in the form of a solution to the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER