A useful convergence fact.

January 17th, 2007 ~ Posted in: Mathematics

If z_n \rightarrow L then (z_1 z_2 \cdots z_n)^{\frac{1}{n}} \rightarrow L also. Attempt a proof, using as elementary methods as you can.

Now use this fact to show that \frac{1}{n} (n!)^{\frac{1}{n}} \rightarrow \frac{1}{e}, given the definition e = \lim_n \left( 1 + \frac{1}{n}\right)^n.

I’ll post my answer when I complete it later tonight.

Mathematics ~

7 Responses to “A useful convergence fact.”

  • 1. AnonEcon
    January 17th, 2007 at 11:21 pm

    –SPOILER ALERT: PROPOSED SOLUTION FOR 2nd PART BELOW–

    Let y_n = n!/n^n

    Let z_1 = y_1
    and z_{n+1} = y_{n+1}/y_n = {1 \over (1+1/n)^n}

    Then z_n \rightarrow 1/e
    [by the definition of e in the problem]

    (y_n)^{1/n} = (\prod_1^n z_n)^{1/n} \rightarrow 1/e [by the first part]
    ————————————
    For the first part, I was thinking of emulating the proof of the root test.

  • 2. AnonEcon
    January 18th, 2007 at 12:26 am

    Sorry, the software is eating up my comments again.

    By using the arithmetic mean-geometric mean inequality we have

    {n \over {\sum (1/z_n)}} \leq (\prod z_n)^{1/n} \leq {{\sum z_n} \over n}

    Both the expression on the extreme left and the extreme right tend to L (the former a bit more work to show when L=0), whence the result follows by the sandwich theorem.

    This argument only works when z_n > 0 but I guess that the theorem too is true and meaningful only in that case.

  • 3. Alex
    January 18th, 2007 at 11:14 am

    Nice. Can you prove the first statement by going all the way back to an \epsilon \delta argument?

  • 4. AnonEcon
    January 18th, 2007 at 8:29 pm

    Apologies, moderator. Another experiment, this time with TeX: \epsilon < 0. It would be nice if we could have a comment preview option.

  • 5. AnonEcon
    January 18th, 2007 at 8:29 pm

    If you mean applying \epsilon\delta to the expression in the middle, then I have not been able to do that so far. Do you have something that works?

    But I can give detailed proofs that the sums on the left and the right converge to L

    First, a small correction. In the inequality above I misused n both as an index and as a separate variable. I should have written,
    {n \over {\sum_1^n{1/z_i}}} \leq \left( \prod_1^n z_i \right)^{1/n} \leq {{\sum_1^n z_i} \over n}

    To prove that sum on the right converges to L: for any \delta>0, let us choose 0<\epsilon<\delta. Since x_i \to L, there is a N such that for i>N, |x_i-L|<\epsilon

    Now,
    \left| {{\sum_1^n x_i}\over n} - L \right|
    \leq {{\sum_1^n {|x_i-L|}}\over n}
    For n>N this equals
    {\sum_1^N {|x_i-L|}+\sum_N^n {|x_i-L|}}\over n
    \leq {{\sum_1^N {|x_i-L|} + (n-N)\epsilon}\over n}
    = \left( \sum_1^N {|x_i-L|} \right)/n + (1-N/n)\epsilon

    As n \to \infty, the expression above tends to \epsilon and hence becomes smaller than \delta. Since we chose \delta as an arbitrary positive number, our result is proved.

    For the sum on the left-hand side we can apply the result above on the series 1/x_i if L>0. The case L=0, I proved with some more \epsilon pushing, which is left as an exercise for the reader.

  • 6. AnonEcon
    January 18th, 2007 at 10:24 pm

    Finally I see the light: catch x_i ultimately between L(1+\epsilon) and L/(1+\epsilon) and use “y^{1/n} \to 1 for all y>0” to make the initial ugliness disappear. You have a direct proof.

    Back for a while I was feeling like a real grown-up analyst, using inequalities and all.

  • 7. Alex
    January 19th, 2007 at 4:56 pm

    I’ll look into getting a comment preview setup.
    “use y^{1/n}\rightarrow 1”, yep, that’s the idea I’m trying to work with, but it’s tricky balancing your n and \epsilon so that the initial terms which aren’t close to L go to 1 and the remaining terms which are really close to L stay close in geometric mean.

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