Geometric Sequence and the Limiting Value

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The discussion focuses on the calculations of a sequence defined by the recurrence relation un+1 = 3 - 1/3(un). The calculations show that the terms converge to a limiting value of 9/4 as n approaches infinity. It is clarified that the sequence is not geometric because it does not satisfy the geometric sequence condition of un+1 = λ un. Participants also discuss formatting options for better readability in their posts. The conclusion confirms the correctness of the limiting value calculation.
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Homework Statement
Hello, I have been practising arithmetic and geometric sequences when I came across the problem below but I am not sure whether I have arrived at the correct conclusion, since the values I have determined for the first few terms in the sequence do not have a constant common ratio as defined by un+1/un=r. Have I made a mistake in my calculations?

1. Find the values of u2,u3 and u4 given tht un+1=3-1/3(un) and u1=3
2. Find the limiting value of un as n tends to infinity
Relevant Equations
un+1=3-1/3(un)
1. When n=1,
u1+1=3-1/3(u1)
u2=3-1/3(3)
u2=2

When n=2
u2+1=3-1/3(u2)
u3=3-1/3(2)
u3=7/3

When n=3
u3+1=3-1/3(u3)
u4=3-1/3(7/3)
u4=20/9

The common ratio is defiend by r=un+1/un, but this is different between the terms, i.e. u2/u1=2/3 whereas u3/u2=(7/3)/2=7/6
Have I made a mistake?

2. A sequence u1, u2, u3... converges to a limit L as the terms get ever closer to L. If the limit of un as n → ∞ is L, the terms un and un+1 are approximately equal to L.
un+1=3-1/3(un)
Replace un+1 and un with L in the equation.
L=3-1/3L
4/3L=3
4L=9
L=9/4

Thus, the limiting value of the sequence is 9/4. Would this be correct?
 
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Your calculations are correct.

The sequence is not geometric; if it were it would satisfy u_{n+1} = \lambda u_n rather than u_{n+1} = \lambda u_n + c.
 
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In the top toolbar there should be a ...∨ sign which allows you to write subscipts.
 
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pasmith said:
Your calculations are correct.

The sequence is not geometric; if it were it would satisfy u_{n+1} = \lambda u_n rather than u_{n+1} = \lambda u_n + c.
Thank you for your reply. Oh ok thank you for the clarification, would my solution for the limiting value also be correct? 😊
 
epenguin said:
In the top toolbar there should be a ...∨ sign which allows you to write subscipts.
I have just looked in the toolbar and have seen what you mean, in addition to the option to insert symbols. Thank you for the suggestion, that is so useful and will definitely allow me to type in a more readable format 👍
 

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