Comments on: A useful convergence fact. http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/ somewhere near the beginning. Mon, 01 Dec 2008 22:27:05 +0000 http://wordpress.org/?v=2.5 By: Alex http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-84374 Alex Fri, 19 Jan 2007 23:56:16 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-84374 I'll look into getting a comment preview setup. "use [tex]y^{1/n}\rightarrow 1[/tex]", yep, that's the idea I'm trying to work with, but it's tricky balancing your [tex]n[/tex] and [tex]\epsilon[/tex] 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. 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.

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-84101 AnonEcon Fri, 19 Jan 2007 05:24:41 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-84101 Finally I see the light: catch [tex]x_i[/tex] ultimately between [tex]L(1+\epsilon)[/tex] and [tex]L/(1+\epsilon)[/tex] and use "[tex]y^{1/n} \to 1[/tex] for all [tex]y>0[/tex]" 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. 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.

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-84082 AnonEcon Fri, 19 Jan 2007 03:29:59 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-84082 If you mean applying [tex]\epsilon\delta[/tex] 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 [tex]L[/tex] First, a small correction. In the inequality above I misused [tex]n[/tex] both as an index and as a separate variable. I should have written, [tex] {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} [/tex] To prove that sum on the right converges to L: for any [tex]\delta>0[/tex], let us choose [tex]0<\epsilon<\delta[/tex]. Since [tex]x_i \to L[/tex], there is a [tex]N[/tex] such that for [tex]i>N[/tex], [tex]|x_i-L|<\epsilon[/tex] Now, [tex]\left| {{\sum_1^n x_i}\over n} - L \right|[/tex] [tex]\leq {{\sum_1^n {|x_i-L|}}\over n}[/tex] For [tex]n>N[/tex] this equals [tex]{\sum_1^N {|x_i-L|}+\sum_N^n {|x_i-L|}}\over n[/tex] [tex]\leq {{\sum_1^N {|x_i-L|} + (n-N)\epsilon}\over n} [/tex] [tex] = \left( \sum_1^N {|x_i-L|} \right)/n + (1-N/n)\epsilon[/tex] As [tex]n \to \infty[/tex], the expression above tends to [tex]\epsilon[/tex] and hence becomes smaller than [tex]\delta[/tex]. Since we chose [tex]\delta[/tex] 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 [tex]1/x_i[/tex] if [tex]L>0[/tex]. The case [tex]L=0[/tex], I proved with some more [tex]\epsilon[/tex] pushing, which is left as an exercise for the reader. 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&gt;0, let us choose 0&lt;\epsilon&lt;\delta. Since x_i \to L, there is a N such that for i&gt;N, |x_i-L|&lt;\epsilon

Now,
\left| {{\sum_1^n x_i}\over n} - L \right|
\leq {{\sum_1^n {|x_i-L|}}\over n}
For n&gt;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&gt;0. The case L=0, I proved with some more \epsilon pushing, which is left as an exercise for the reader.

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-84081 AnonEcon Fri, 19 Jan 2007 03:29:04 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-84081 Apologies, moderator. Another experiment, this time with TeX: [tex] \epsilon < 0[/tex]. It would be nice if we could have a comment preview option. Apologies, moderator. Another experiment, this time with TeX: \epsilon &lt; 0. It would be nice if we could have a comment preview option.

]]> By: Alex http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-83923 Alex Thu, 18 Jan 2007 18:14:36 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-83923 Nice. Can you prove the first statement by going all the way back to an [tex]\epsilon \delta[/tex] argument? Nice. Can you prove the first statement by going all the way back to an \epsilon \delta argument?

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-83801 AnonEcon Thu, 18 Jan 2007 07:26:15 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-83801 Sorry, the software is eating up my comments again. By using the <a href="http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means" rel="nofollow">arithmetic mean-geometric mean inequality</a> we have [tex] {n \over {\sum (1/z_n)}} \leq (\prod z_n)^{1/n} \leq {{\sum z_n} \over n} [/tex] Both the expression on the extreme left and the extreme right tend to [tex]L[/tex] (the former a bit more work to show when [tex]L=0[/tex]), whence the result follows by the sandwich theorem. This argument only works when [tex]z_n > 0[/tex] but I guess that the theorem too is true and meaningful only in that case. 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.

]]> By: AnonEcon http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact/#comment-83779 AnonEcon Thu, 18 Jan 2007 06:21:19 +0000 http://www.tangentspace.net/cz/archives/2007/01/a-useful-convergence-fact#comment-83779 --SPOILER ALERT: PROPOSED SOLUTION FOR 2nd PART BELOW-- Let [tex]y_n = n!/n^n[/tex] Let [tex]z_1 = y_1[/tex] and [tex]z_{n+1} = y_{n+1}/y_n = {1 \over (1+1/n)^n}[/tex] Then [tex]z_n \rightarrow 1/e[/tex] [by the definition of [tex]e[/tex] in the problem] [tex] (y_n)^{1/n} = (\prod_1^n z_n)^{1/n} \rightarrow 1/e[/tex] [by the first part] ------------------------------------ For the first part, I was thinking of emulating the proof of the root test. –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.

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