ChapterZero

A useful convergence fact.

Filed under: Mathematics — Alex @ 6:34 pm 1/17/2007

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.

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

Let

Let
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 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.

Comment by AnonEcon — 1/17/2007 @ 11:21 pm
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 (the former a bit more work to show when ), whence the result follows by the sandwich theorem.

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

Comment by AnonEcon — 1/18/2007 @ 12:26 am
Nice. Can you prove the first statement by going all the way back to an $\epsilon \delta$ argument?

Comment by Alex — 1/18/2007 @ 11:14 am
Apologies, moderator. Another experiment, this time with TeX: $\epsilon < 0$ . It would be nice if we could have a comment preview option.

Comment by AnonEcon — 1/18/2007 @ 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

First, a small correction. In the inequality above I misused 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 such that for , $|x_i-L|<\epsilon$

Now,
$\left| {{\sum_1^n x_i}\over n} - L \right|$
$\leq {{\sum_1^n {|x_i-L|}}\over n}$
For 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 if . The case , I proved with some more $\epsilon$ pushing, which is left as an exercise for the reader.

Comment by AnonEcon — 1/18/2007 @ 8:29 pm
Finally I see the light: catch ultimately between $L(1+\epsilon)$ and $L/(1+\epsilon)$ and use “ $y^{1/n} \to 1$ for all ” 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.

Comment by AnonEcon — 1/18/2007 @ 10:24 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 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.

Comment by Alex — 1/19/2007 @ 4:56 pm

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A useful convergence fact.

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