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Prove that |z|^n -> 0 if |z| < 0

  1. Aug 1, 2011 #1
    Prove that |z|^n ---> 0 if |z| < 0

    1. The problem statement, all variables and given/known data

    This isn't really a homework problem; I'm preparing for an upcoming course and trying to recall how to do these basic proofs.

    2. Relevant equations

    Definition of convergence

    3. The attempt at a solution

    We seek a positive integer N such that for any positive number ∂,
    | |z|n - 0 | < ∂ whenever n ≥ N.

    Fix ∂ > 0.

    | |z|n- 0| = |z|n < ∂
    n < log(∂) / log(|z|) (some steps omitted)

    But here's the problem: I have n smaller than a positive number, since log(∂), log(|z|) < 0 when ∂, |z| < 1. What do?
  2. jcsd
  3. Aug 1, 2011 #2
    Re: Prove that |z|^n ---> 0 if |z| < 0

    I'd split it up into cases just to be a bit cleaner:

    Let [itex]\epsilon > 0[/itex] be given and suppose [itex]|z| < 1[/itex].

    If [itex]\epsilon \geq 1[/itex], then [itex]|z|^n < 1 \leq \epsilon[/itex] for what values of n?

    If [itex]\epsilon < 1[/itex], put [itex]|z| = 1/x[/itex]. Then what must be true of x?
  4. Aug 1, 2011 #3


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    Re: Prove that |z|^n ---> 0 if |z| < 0

    Note that stringy is assuming that you mean |z|< 1, not |z|< 0 which is impossible anyway.
  5. Aug 1, 2011 #4
    Re: Prove that |z|^n ---> 0 if |z| < 0

    n > log(eps) / log(|z|)

    Then x > 1 (?????)
  6. Aug 1, 2011 #5
    Re: Prove that |z|^n ---> 0 if |z| < 0

    Certainly [itex]\left|z\right|^n < 1[/itex] for all n! Remember, we're assuming the absolute value of z is less than 1.

    EDIT: This is the sort of trivial case that is dispensed with quickly at the beginning of the proof.

    Bingo. Then proving [itex] \left|z\right|^n \to 0[/itex] is the same as proving [itex]1/x^n \to 0 [/itex] with x>1.
    Last edited: Aug 1, 2011
  7. Aug 1, 2011 #6
    Re: Prove that |z|^n ---> 0 if |z| < 0

    Thanks, brah!

    I have another one: |z|n / n! ----> 0

    How should I go about showing this?

    |z|n / n! < ∂
    log( |z|n / n! ) < log(∂)
    log(|z|n) - log(n!) < log(∂)
    n * log(|z|) - ∑log(k) < log(∂) (where the index of the sum is k = 2, 3, ..., n)

    ?????? Seems like I'm getting nowhere with that

    |z|n / n! < ∂
    [∑n!xn-k(iy)k/(n-k)!k!] / n! < ∂
    [∑xn-k(iy)k/(n-k)!k!] < ∂

    ???? Doesn't seem legit
  8. Aug 1, 2011 #7
    Re: Prove that |z|^n ---> 0 if |z| < 0

    Hmmm... I have an idea. We are to prove

    [tex] \lim_{n\to \infty} \frac{|z|^n}{n!} = 0,[/tex]

    correct? No conditions placed on the size of |z| (it doesn't matter anyways)? We know

    [tex] \sum_{n=0}^\infty \frac{|z|^n}{n!}[/tex]

    converges. Do you remember why and to what? What do we know about the nth term of a convergent series?

    By the way, remember what n! is:

    [tex] n! = n(n-1)(n-2)...(2)(1). [/tex]

    It's not a sum.
  9. Aug 1, 2011 #8
    Re: Prove that |z|^n ---> 0 if |z| < 0

    I know, but log(n!) = log(n*(n-1)***2*1) = log(n) + log(n-1) + ... + log(2) + log(1)

    The chapter from which this practice problem was derived comes prior to the chapter on series, so I don't think I can use the fact that ∑an is convergent implies an ---> 0. We'll have to go about it another way.
  10. Aug 1, 2011 #9
    Re: Prove that |z|^n ---> 0 if |z| < 0

    D'oh! You're right. I just saw the sigma in there and jumped to conclusions. Let me think about this for awhile. I get back.
  11. Aug 1, 2011 #10
    Re: Prove that |z|^n ---> 0 if |z| < 0

    You remember from calculus the ratio test for series? There's an analogous test for sequences. The proofs are similar.

    Let [itex] (a_n) [/itex] be a sequence of nonzero terms such that

    [tex] \lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n}\right| =\alpha .[/tex]

    If [itex]\alpha< 1[/itex], then the sequence [itex](a_n)[/itex] goes to zero.

    Quick sketch of a proof: Suppose [itex]| a_{n+1}/a_n| \to \alpha [/itex] and [itex] \alpha <1[/itex]. Then there is some N so that n>N implies [itex] |a_{n+1}/a_n| \leq \beta[/itex] where [itex] \alpha < \beta < 1 [/itex]. So if n>N,

    [tex] |a_n| = \frac{|a_n|}{|a_{n-1}|} \frac{|a_{n-1}|}{|a_{n-2}|} \cdots \frac{|a_{N+1}|}{|a_{N}|} |a_{N}| \leq \beta^{n-N} |a_N| = \beta^n \frac{|a_N|}{\beta^N}.[/tex]

    Since [itex]\beta <1[/itex], we have [itex]\beta^n \to0 [/itex] (we just proved this!). Therefore, by the squeeze theorem, [itex] (|a_n|)[/itex] and hence [itex](a_n)[/itex] goes to zero.

    Apply this test to your sequence and you'll have your desired result.

    You can also obtain the result by using known upper and lower bounds on the factorial. You can look them up on the web. Trying to prove that the sequence converges by explicitly finding a delta is going to be tough, which is why I've avoided that method of attack.
  12. Aug 2, 2011 #11
    Re: Prove that |z|^n ---> 0 if |z| < 0

    If I might 'hijack' this question, I have the same question, but it explicitly states to avoid logarithms and exponentials, and use the properties of convergence and the properties of a subsequence u2n to prove un = |a|^n ---> 0 when n ---> infinity if |a| < 1. The hint supplied is to let u2n = (un)^2. I'm assuming this makes use of the fact that all subsequences converge to the same limit as the sequence? And also, is application of the Bolzano-Weierstrass theorem required for this proof?
  13. Aug 2, 2011 #12
    Re: Prove that |z|^n ---> 0 if |z| < 0

    If you want to avoid logarithms, you can use Bernoulli's inequality which states

    [tex](1+b)^n \geq 1+nx[/tex]

    for all positive integers [itex]n[/itex] and real [itex]b\geq -1[/itex].

    Since [itex] |a| <1[/itex], write

    [tex] |a| = \frac{1}{1+b} [/tex]

    where [itex]b>0[/itex]. Apply Bernoulli's inequality and you'll find a delta relatively easily.

    As for this subsequence business... I'm not immediately seeing the reason why we should look at that subsequence! :confused:

    We know that [itex](u_n)[/itex] converges to a point u if and only if all subsequences converge to u. So it's not enough to ONLY show that [itex](u_{2n})[/itex] goes to zero. We'd need to show all subsequential limits are 0.

    However, we know that monotonic sequences converge if and only if they are bounded. We know [itex](u_n)[/itex] is bounded (below by 0 and above by 1). So we know [itex](u_n)[/itex] converges to a point u. Then if you show [itex](u_{2n})[/itex] goes to zero, you know u=0.

    Hmmmm... I'll think about this one for awhile and get back if I come up with anything. Meanwhile, maybe somebody else can chime in.
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