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Sequence convergence

  • Thread starter estro
  • Start date
  • #1
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I know that the sequences meets the following:
[tex](n+1)(a_{n+1}-a_n)=n(a_{n-1}-a_n)[/tex]
I've got the feeling that this sequence is alternating or decreasing, but I was unable to prove it.
Usually I use induction to prove things about such inductive sequence but in this case I don't have real values t o the first terms of the sequence so I don't have an idea how to even begin inductive prove.
 
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Answers and Replies

  • #2
tiny-tim
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hi estro! :smile:

(try using the X2 icon just above the Reply box :wink:)
(n+1)(a_{n+1}-a_n)=n(a_{n+1}-a_n)
are those two brackets meant to be the same? :confused:

if (n+1)X = nX, then X = 0
 
  • #3
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I have found this question in an old exam, so maybe there is some kind of mistake, I'm not sure.
[tex]
(n+1)(a_{n+1}-a_n)=n(a_{n-1}-a_n)
[/tex]
[EDIT] Copied question in a wrong way, now fixed.
 
  • #4
tiny-tim
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ah! :rolleyes:

ok … standard way of solving that is to define bn = an+1 - an wink:
 
  • #5
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Still can't see a benefit to define [tex]b_m=a_{n+1}-a_n[/tex].
I've used to prove convergence of alternating sequences by using Cantor's Lemma in a geometric formulation.
Marked each section [tex][a_{n+1},a_n][/tex] and then I proved that each next section smaller and ultimately the length of a sections is going to 0.
But I always had the first terms and used induction to prove sequences is bounded.
But here all these tricks don't work as I can't even prove that this sequence is alternating.
 
  • #6
Gib Z
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If you let [itex] b_n = a_n - a_{n-1} [/itex] then your relation is transformed into [tex] \frac{b_{n+1}}{b_n} = \frac{n}{n+1} < 1 [/tex].
 

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