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Limits of Sequences

  1. Jan 21, 2005 #1
    Hello all:

    Given [tex] a_n = \frac {n}{\alpha^n} [/tex] and [tex] \alpha [/tex] is a number greater than 1, we assert as n increases the sequence of numbers [tex] a_n = \frac {n}{\alpha^n} [/tex] tends to the limit 0.

    Let us consider the sequence [tex] \sqrt a_n = \frac {\sqrt n}{(\sqrt \alpha)^n}[/tex]

    We put [tex] \sqrt \alpha = 1+h [/tex] where [tex] h > 0 [/tex]

    [tex] \sqrt \alpha^n = (1+h)^n > 1+nh [/tex]

    [tex]\sqrt a_n = \frac {\sqrt n}{(1+h)^n} \leq \frac {\sqrt n}{1+nh} \leq \frac {\sqrt n}{nh} = \frac {1}{h\sqrt n} [/tex]

    [tex] a_n \leq \frac {1}{nh^2} [/tex] and [tex] a_n \rightarrow 0 [/tex]

    Now my question is: Suppose you are asked to find numbers [tex] N_1, N_2, N_3 [/tex] such that

    (a) [tex]\frac {n}{2^n} < \frac {1}{10}[/tex] for every [tex] n > N_1 [/tex]
    (b) [tex]\frac {n}{2^n} < \frac {1}{100}[/tex] for every [tex] n > N_2 [/tex]
    (c) [tex]\frac {n}{2^n} < \frac {1}{1000}[/tex] for every [tex] n > N_3 [/tex]

    Do I have to guess and check for these numbers or can I somehow use the proof of this sequence? Because what if [tex] \epsilon [/tex] becomes smaller and smaller? Then there must be some method to find [tex] N_1, N_2, N_3 [/tex]

    Thanks a lot for any help or advice

    EDIT: Its [tex] \frac {n}{2^n} < \frac {1}{10} [/tex] for every [tex] n > N_1 [/tex] and the same with the others
    Last edited: Jan 21, 2005
  2. jcsd
  3. Jan 21, 2005 #2


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    Homework Helper

    In your case here: [tex]\alpha = 2[/tex] Solve for h.

    You want [tex]a_n<\frac{1}{10}[/tex]. If you know that [tex]\frac{1}{nh^2}<\frac{1}{10}[/tex] then that means that [tex]a_n<\frac{1}{10}[/tex].

    So from [tex]\frac{1}{nh^2}<\frac{1}{10}[/tex] you can get the n value you need. This will give you an N1 such that if n>N1, [tex]a_n<\frac{1}{10}[/tex]

    But are you looking for the minimum N1 such that if n>N1 then [tex]a_n<\frac{1}{10}[/tex], or are you just looking for any N1 such that if n>N1 then [tex]a_n<\frac{1}{10}[/tex]? If you're looking for the minimum N1, then you might want to want to calculate [tex]a_{N1}[/tex] and make sure that it is >=1/10.
  4. Jan 21, 2005 #3
    i am looking for the minimum [tex] N_1 [/tex]

    I know that for part (a) [tex] N_1 = 6 [/tex] buy why?

    Last edited: Jan 21, 2005
  5. Jan 21, 2005 #4


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    Homework Helper

    Then I think you'd just have to plug it in and check.

    Are you trying to do an epsilon-delta proof to show that the limit is 0? Then it is sufficient to show that for every epsilon>0, there is some N such that for n>N |a(n)-0|<epsilon or in other words show that a(n)<epsilon. I'm guess that the 1/10, 1/100 are different choices for epsilon?
  6. Jan 21, 2005 #5
    any other ideas?

  7. Jan 21, 2005 #6
    so i take it that the only way to find the smallest numbers would be by guess and check? Because after solving for n i get 64.

    This is from Courant's book by the way (i am a high school student)

    Last edited: Jan 22, 2005
  8. Jan 22, 2005 #7
    Thanks anyway for your help
  9. Jan 22, 2005 #8
    Same concept Different Limit

    Ok so let's say we have [tex] a_n = \sqrt n+1 - \sqrt n [/tex] and we want to find numbers [tex] N_1, N_2, N_3 [/tex] such that

    [tex] \sqrt n+1 - \sqrt n < \frac {1}{10}[/tex] for every [tex] n > N_1 [/tex]
    [tex] \sqrt n+1 - \sqrt n < \frac {1}{100}[/tex] for every [tex] n > N_2 [/tex]
    [tex] \sqrt n+1 - \sqrt n < \frac {1}{1000}[/tex] for every [tex] n > N_3 [/tex]

    Now since [tex] \sqrt n+1 - \sqrt n = \frac {(\sqrt n+1 - \sqrt n)(\sqrt n+1 + \sqrt n)}{\sqrt n+1 + \sqrt n} = \frac {1}{\sqrt n+1 + \sqrt n}

    How would i obtain [tex] N_1, N_2 ,N_3 [/tex]?

    PS: I know if I solve the equation I get a [tex] N_1 N_2 N_3 [/tex] but I want to get the least of all of them

    Any help is appreciated (as with the first one do I just guess and check?)

  10. Jan 22, 2005 #9
    hmmm, i i think i have to substitute values in lless than the solved value?

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