- #1

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My solution so far is this:

Where have I gone wrong? I'm assuming I'm taking the wrong limit but I'm really stumped here and no idea what else it should be..

I hope my writing is ok, I tried to be as neat as possible!

Thanks

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- Thread starter Firepanda
- Start date

- #1

- 430

- 0

My solution so far is this:

Where have I gone wrong? I'm assuming I'm taking the wrong limit but I'm really stumped here and no idea what else it should be..

I hope my writing is ok, I tried to be as neat as possible!

Thanks

- #2

- 42

- 0

- #3

- 430

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Putting k = n-3 does it really change anything? I can't see what it changes..

Thanks

- #4

- 42

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I'm sorry, it doesn't change a thing. But now I noticed that the sum isn't calculated correctly, you should change k back to n-3 so you can see the flaw. The sum is actually

[itex] \sum_{n=3}^{\infty}\left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)}[/itex] so the cancellations are not correct.

[itex] \sum_{n=3}^{\infty}\left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)}[/itex] so the cancellations are not correct.

Last edited:

- #5

- 430

- 0

I'm sorry, it doesn't change a thing. But now I noticed that the sum isn't calculated correctly, you should change k back to n-3 so you can see the flaw. The sum is actually

[itex]\sum_{n=3}^\infty{\left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)}[/itex] so the cancellations are not correct.

starting at n=3 I get (1 - N

etc

Isn't it the same cancellation?

- #6

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- 0

[itex]a_n= \left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)}[/itex]

[itex]a_{n+1}= \left(\frac{n+1}{n+2}\right)^{a(n-2)}-\left(\frac{n+1}{n+2}\right)^{a(n-1)}[/itex]?

- #7

- 430

- 0

[itex]a_n= \left(\frac{n}{n+1}\right)^{a(n-3)}-\left(\frac{n}{n+1}\right)^{a(n-2)}[/itex]

[itex]a_{n+1}= \left(\frac{n+1}{n+2}\right)^{a(n-2)}-\left(\frac{n+1}{n+2}\right)^{a(n-1)}[/itex]?

Ah of course, woops!

where do I go from here, I assume I can't cancel any of them, right?

Not sure how to sum that series either..

- #8

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cause it looks to me like you should have [itex]f_3^3 = 1/2 \cdot 1 \cdot (1-\frac{3}{3+1})^a[/itex]

[itex]f_3^4 = 1/2 \cdot 1 \cdot (\frac{3}{3+1})^a \cdot (1-\frac{4}{4+1})^a[/itex]

[itex]f_3^5 = 1/2 \cdot 1 \cdot (\frac{3}{3+1})^a \cdot (\frac{4}{4+1})^a \cdot (1-\frac{5}{5+1})^a[/itex]

and so forth. Mayby I'm not interpreting the probabilities correctly, as I said, I'm not familiar with these...

- #9

Ray Vickson

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[tex] \left(\frac{n}{n+1}\right)^a \left(\frac{n+1}{n+2}\right)^a \cdots \left(\frac{n+k-1}{n+k}\right)^a = \left( \frac{n}{n+k}\right)^a, [/tex]

which --> 0 as k --> infinity. You can fix up this argument to show that [itex] \Pr \{ X_{n+k}=5 \; \mbox{ i.o. } |X_n = 5 \} = 0, [/itex], so state 5 is left with probability 1 at a finite time. When that happens, the system returns to state 3, implying that state 3 is recurrent.

RGV

- #10

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So first we want to find out what [itex]f_3^n[/itex] is when [itex]n\geq 3:[/itex]

[itex]f_3^n = \displaystyle \frac{1}{2}\cdot1\cdot\left(\prod_{k=3}^{n-1}\left(\frac{k}{k+1}\right)^a\right)\cdot\left(1-\left(\frac{n}{n+1}\right)^a\right) = \frac{1}{2}\cdot\left(\frac{(n-1)!}{2}\cdot\frac{2\cdot3}{n!}\right)^a\cdot\left(1-\left(\frac{n}{n+1}\right)^a\right)[/itex]

[itex] \displaystyle = \frac{1}{2} \cdot \left( \frac{3}{n} \right)^a \cdot \left(1-\left( \frac{n}{n+1} \right)^a \right) = \frac{3^a}{2} \cdot \left(\frac{1}{n^a}- \frac{1}{(n+1)^a}\right)[/itex]

So now the sum of the probabilities is easy to calculate, and I believe that you will finally get the correct answer.

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