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Sequence satisfying a condition for all n

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that a sequence {s_n} of positive numbers satisfies the condition s_(n+1) > αs_n for all n where α > 1. Show that s_n → ∞

    My teacher mentioned something about making it into a geometric sequence and taking the log. I'm just confused.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 22, 2009 #2

    LCKurtz

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    You can start with s2 > as1. Now what about s3? Can you compare it to s2 and s1? Continue...
     
  4. Sep 22, 2009 #3

    HallsofIvy

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    So [itex]s_2> a s_n[/itex], [itex]s_3> a s_2> a(a s_1)= a^2 s_1[/itex], [itex]s_4> a s_3> a(a^2 s_1)= a^3 s_1[/itex]. So [itex]s_n> [/itex] a to what power times [itex]s_1[/itex]? What does that have to do with a "geometric sequence"?
     
  5. Sep 22, 2009 #4
    If [tex](s_n)[/tex] is a sequence and the limit [tex]\lim_{n \to \infty}|s_{n+1} / {s_n}| = L [/tex] exists and [tex]L < 1[/tex], then [tex]\lim s_n[/tex] converges. If not, what do you think happens?
     
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