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May 19th, 2017, 06:40 AM   #1
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Recurrence Sequence

The non-zero values for $x_0$ and $x_1$ such that the sequence defined by the recurrence relation $x_{n+2} = 2x_n $, is convergent are
A) $x_0 = 1$ and $x_1 = 1$
B) $\displaystyle x_0 = \frac{1}{2}$ and $\displaystyle x_1 = \frac{1}{4}$
C) $\displaystyle x_0 = \frac{1}{10}$ and $\displaystyle x_1 = \frac{1}{20} $
D) none of the above

I have checked the sequence, but I was not able to understand which is monotonically increasing or decreasing to check the convergence.
Option A is having the sequence like 1,1,2,2,4,4,8,8... Is it convergent?
Please help!

Last edited by skipjack; May 19th, 2017 at 09:01 AM.
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May 19th, 2017, 06:50 AM   #2
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I would calculate numbers for each of A, B, C and D first. If you need more information after that, look at the definition of convergence: informally a sequence is convergent if it gets closer and closer to some limiting value. Is that true of 1,1,2,2,4,4,8,8?

An alternative approach is to consider that every subsequence of a convergent sequence converges to the same limit as the sequence. So, does the subsequence $(x_0, x_2, x_4, \ldots)$ converge?

Last edited by v8archie; May 19th, 2017 at 06:52 AM.
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May 19th, 2017, 07:26 AM   #3
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Quote:
Originally Posted by v8archie View Post
I would calculate numbers for each of A, B, C and D first. If you need more information after that, look at the definition of convergence: informally a sequence is convergent if it gets closer and closer to some limiting value. Is that true of 1,1,2,2,4,4,8,8?

An alternative approach is to consider that every subsequence of a convergent sequence converges to the same limit as the sequence. So, does the subsequence $(x_0, x_2, x_4, \ldots)$ converge?
The subsequence $2^n$ is divergent in Ratio test. Does it mean the original sequence also diverges?

All subsequences of the 3 options are diverging... Does it mean none of the above is the correct option?

Last edited by skipjack; May 19th, 2017 at 09:02 AM.
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May 19th, 2017, 09:02 AM   #4
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Yes.
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May 19th, 2017, 09:04 AM   #5
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Yes.
You are telling YES to my last message or to my first question ??
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May 19th, 2017, 09:28 AM   #6
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Both. Think about it.
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May 19th, 2017, 09:35 AM   #7
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Quote:
Originally Posted by v8archie View Post
Both. Think about it.
If a subsequence is divergent then it's original sequence also diverges ?
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