Talk:Divergent series

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[edit] 1 − 1 + 1 − 1 + · · ·

For example, Cesàro summation assigns the divergent series

1 - 1 + 1 - 1 + \cdots

the value 1 \over 2.

Well, no, the average of the sum of N terms → 0, evidently. But perhaps some other example was meant. Such as

1 + 0 + 1 + 0 + ... .

Charles Matthews 16:20, 7 Sep 2004 (UTC)

Not so: the average of PARTIAL sums is important, no? and the average of 1, 0, 1, 0, etc. is? —The preceding unsigned comment was added by Danwbartlett (talk • contribs) 00:00, 11 March 2005 (UTC)
Charles, It depends what summation method you're using.
S = (1-1)+(1-1)+(1-1)... gives us S=0 , but..
S = 1+(-1+1)+(-1+1)+(-1+1).... gives us S=1 , and then again..
S = 1-(1-1+1-1+1-1...) gives us S=1-S, and thus 2S=1, and therefore S=1/2.
so I guess the third example must known as Cesaro Summation as described above. —The preceding unsigned comment was added by 203.59.206.158 (talk • contribs) 09:33, 28 January 2007 (UTC)
For sums of this series, please refer to Summation of Grandi's series or related articles. Melchoir 19:30, 29 January 2007 (UTC)

[edit] Removal

Cut this out, for a couple of reasons.

"== Proving an infinite series is divergent ==

There is a quick test to prove an infinite series is convergent or divergent.

Given an infinite series, if

\lim_{n \to \infty} \sum_{n=1}^\infty a_k \ne 0

then ak is said to be divergent.

[edit] Example

Prove whether the following is convergent or divergent.

\lim_{n \to \infty} \sum_{n=1}^\infty cos \frac{1}{k^2}

Solution:

\lim_{n \to \infty} cos \frac{1}{k^2} \to 1

1 \ne 0

Therefore, this series is divergent."

Reason (1): the test is mis-stated. A necessary condition for convergence is that the ak → 0. Reason (2) is that this really belongs on the basic infinite series page. Charles Matthews 06:19, 12 Feb 2005 (UTC)

This is the first of the convergence criteria given at infinite series, in fact. Charles Matthews 06:21, 12 Feb 2005 (UTC)

Oops, I looked there. I need to look more carefully. Tygar 06:49, Feb 12, 2005 (UTC)

[edit] Second sentence?

The second sentence of this article doesn't seem to add anything. (No one is reading this in order to learn what "antonym" means.) But I wanted to see if there are objections before deleting it. (I would probably put the link to "convergent series" on the word "converge" at the end of the first sentence.) Dchudz 23:26, 6 November 2006 (UTC)


[edit] issue with the Hahn-Banach paragraph

"The operator giving the sum of a convergent sequence is linear, and it follows from the Hahn-Banach theorem that it may be extended to a summation method summing any bounded sequence."

We talk about summing series, and not sequences, right? And isn't it the sequence of partial sums that needs to be bounded, and not terms of the series? If there aren't objections, I'll change it to:

"The operator giving the limit of a convergent sequence is linear, and it follows from the Hahn-Banach theorem that it may be extended to a summation method summing any series whose sequence of partial sums is bounded."

(Or anyone who is confident that I'm right could just go ahead and make the change.)

Dchudz 23:34, 6 November 2006 (UTC)

I think it's okay to speak about summing a sequence. However, a convergent sequence is something else as a convergent series, so you're right that it should be changed. I replaced it by "The operator giving the sum of a convergent series is linear, and it follows from the Hahn-Banach theorem that it may be extended to a summation method summing any series with bounded partial sums." which is a slight reformulation of what you proposed. -- Jitse Niesen (talk) 01:07, 7 November 2006 (UTC)
Yeah, that's good. Thanks. 130.58.219.209 02:54, 7 November 2006 (UTC)

Is there a way to do the Hahn-Banach extension so that your summation method ends up having "stability" (regularity and linearity are clear...)?Dchudz 18:20, 14 December 2006 (UTC)

[edit] Definition of "converge"

The second sentence states, "if a series converges, the individual terms of the series must approach zero," but this contradicts the definition given on the infinite series page, which states, "this limit can have a finite value; if it does, the series is said to converge..."

Obviously, there are finite values other than zero. Am I missing something? Phlake 10:29, 12 November 2006 (UTC)

The first sentence you quote says:
"if the series a_1 + a_2 + a_3 + \cdots converges, then \lim_{i\to\infty} a_i = 0."
The second sentence you quote says:
"the series a_1 + a_2 + a_3 + \cdots converges if the limit \lim_{n\to\infty} \sum_{i=1}^n a_i has a finite value."
The first case is about the limit of the individual terms. The second case is about the limit of partial sums. I hope that answers your question. -- Jitse Niesen (talk) 11:07, 12 November 2006 (UTC)
Aha. That's what I missed. Thanks. Phlake 13:40, 19 November 2006 (UTC)
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