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|a| < 1 show lim |a|^n ->0

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data

    given |a| < 1, show that the limit of |a|^n goes to 0 as n goes to infinity.

    2. Relevant equations



    3. The attempt at a solution

    let |a|<1 and n>0 (n is a natural number, a is a real number)

    then

    |a^n| < 1^n

    then

    |a|^n < 1

    then

    1/n * |a|^n <= |a|^n < 1 (and we know from earlier that 1/n * |a|^n goes to zero)

    so 0 <=|a|^n < 1 meaning |a|^n is bounded below.

    now note that

    |a| > |a|^2 > |a|^3 > ... |a|^n > |a|^(n+1) > ... since a < 1

    Since a^n is progressively smaller and bounded below by zero, we know that a^n does converge by the convergent monotone sequence thm and since there are infinite items in this sequence, its limit cannot be in the set of a^n.

    therefore let L be the limit,

    |a^n - L| < e, since L is smaller than a^n;

    a^n - L < e.

    **I tried using the squeeze theorem but could not find a sequence that I new to be greater than a^n and goes to zero.** And I don't know what I am missing in order to be able to claim that L is zero.
     
  2. jcsd
  3. Nov 26, 2013 #2

    jbunniii

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    Hint: ##|a|^{n+1} = |a| \cdot |a|^{n}##. What happens if you take limits of both sides of this equation?
     
  4. Nov 26, 2013 #3
    i am still missing something, cause now i am getting the lim to be e which i know is incorrect.

    (n+1) * lim ln |a|= Lim (|a| * |a|^n) = lim |a| * n lim ln |a|

    (n+1) = n lim |a|

    (1+1/n) = lim |a|

    (1+1/n)^n = (lim |a|)^n

    e = lim |a|^n
     
  5. Nov 26, 2013 #4

    jbunniii

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    I'm not sure why you're taking logs, but in any case the above doesn't make any sense.

    First, how did you slide the "lim" past the ##n+1## to get ##(n+1) \lim \ln |a|##? Aren't you taking the limit with respect to ##n##?

    Second, it is not true that ##\ln (|a| \cdot |a|^n) = |a| \cdot n\ln|a|## as your second equality seems to suggest.
     
  6. Nov 26, 2013 #5

    jbunniii

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    Try taking limits directly of the left and right hand sides. Don't take logs first:
    $$|a|^{n+1} = |a| \cdot |a|^n$$
    Take advantage of the fact that you already concluded that ##\lim |a|^n## must exist because the sequence is bounded and monotonic.
     
  7. Nov 26, 2013 #6
    I am unsure. I remember from an earlier class that the limit of x^n is equal to n * lim ln x

    so i tried extending it to calculate the limits like you suggested:

    lim of a^(n+1) = (n+1) * lim ln a [1]

    lim (a*a^n) = lim a * lim a^n = lim a * n * lim ln a. [2] Then i divided both sides ([1], [2]) by lim ln a
     
  8. Nov 26, 2013 #7

    jbunniii

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    It's not true. What is true is that ##\ln (x^n) = n \ln (x)## and therefore ##\lim \ln (x^n) = \lim n \ln (x)##, but that isn't going to help you here.

    You have established that ##\lim_{n\rightarrow \infty} |a|^n## must exist. Let's call this limit ##L##. Now what is ##\lim_{n\rightarrow \infty} |a|^{n+1}##?
     
  9. Nov 26, 2013 #8
    lim |a|^(n+1) = aL < L since a < 1
     
  10. Nov 26, 2013 #9

    jbunniii

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    What else is ##\lim|a|^{n+1}## equal to? Think about sequences in general. If ##(x_n)## is some sequence and ##\lim x_n = L##, then what is ##\lim x_{n+1}##?
     
  11. Nov 26, 2013 #10
    It should have the same limit
     
  12. Nov 26, 2013 #11

    jbunniii

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    Yes, that's right. So if we put ##x_n = |a|^n##, then ##x_{n+1} = |a|^{n+1}##.

    If ##\lim |a|^n = L##, then what is ##\lim |a|^{n+1}##?
     
  13. Nov 26, 2013 #12
    lim |a|^(n+1) = L
     
  14. Nov 26, 2013 #13
    but doesn't that show

    lim |a|^(n+1) = lim |a| * lim |a|^n

    L = lim |a| * L
    1 = lim |a| ??
     
  15. Nov 26, 2013 #14

    jbunniii

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    Yes!

    Well, that would be valid if ##L## was nonzero. Notice that ##\lim |a| = |a|##, so what you have just shown is that if you can divide by ##L##, then ##|a| = 1##. But we are given that ##|a| < 1##.

    We want to prove that ##L## is zero, so dividing by ##L## is not what you want to do. How can you solve ##L = \lim |a| \cdot L## for ##L##?
     
  16. Nov 26, 2013 #15
    since |a| < 1 then lim |a| = |a| < 1 and L = lim |a| * L then L - L*Lim |a| = 0
    which implies L(1-Lim|a|)=0 therefore L = 0
     
  17. Nov 26, 2013 #16
    got it!! thanks!
     
  18. Nov 26, 2013 #17

    jbunniii

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    Looks good.
     
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