Talk:Telescoping series

message: Not all telescoping sums are infinite series; some are finite.  Should this be moved to telescoping sum?  (For a nice example, see the section on probability distributions at order statistic.) Michael Hardy 01:07, 9 Feb 2005 (UTC)
replies: text: While I agree that the title is misleading, a quick websearch notes that many math educational sites like Khan Academy and Brilliant refer to them as Telescoping Series. However, the esteemed Wolfram Alpha refers to them as Telescoping Sums. For this reason, I will add the terminology to the page. Fantasticawesome (talk) 21:06, 19 September 2019 (UTC)
thread_title:  Change Title to Telescoping Sums? 

message: "Example 1" is essentially the same as the earlier example. Michael Hardy 22:33, 3 December 2005 (UTC)
replies: text: I propose putting the second occurance first, then putting something like, "This is useful to prove convergence, because as N  \rightarrow \infty the sum tends to 1."  Then removing that occurance.  x42bn6 Talk 07:27, 4 December 2005 (UTC)
thread_title: Why use essentially the same example twice?

message: I have found some stuff about this article, and I am rewriting it now at User:x42bn6/Working On/Telescoping series.  I would appreciate it if nobody did any major edits to this article, but I welcome feedback at User_talk:x42bn6/Working On/Telescoping series.  x42bn6 Talk 13:24, 9 December 2005 (UTC)
thread_title:  Rewrite 

message: Is there a possibility of a proof that relates for telescoping series? By that I mean, is there a formula for finding out what the final value is when it converges? If anyone really wants to talk about it, you can discuss it here. —EdBoy 13:10, 14 August 2006 (UTC)
replies: text: You may want to ask Wikipedia:Reference Desk/Mathematics but they will probably tell you to do it yourself, using limits.  x42bn6 Talk 02:11, 15 August 2006 (UTC)
thread_title: Proof?

message: I think it would be to include some practical applications of telescoping series in this article, but so far I have not found any. —The preceding unsigned comment was added by Panchaos (talk • contribs) 16:08, 14 March 2007 (UTC)
replies: text: I've added an application.  I think others can be found by clicking on "what links here". Michael Hardy 20:50, 4 September 2007 (UTC)
thread_title:  Applications? 

message: The pitfalls section is bad as it stands. The argument used for the valid sum could be used for the invalid example. It is necessary to use a convergence criterion that distinguises between them. 
It is easy to create an example where the terms and the grouped terms tend to zero which does not converge. Create a series as follows: 1 - 1 + 1/2 + 1/2 - 1/2 - 1/2, + 1/4 + 1/4 + 1/4 + 1/4 - 1/4 - 1/4 - 1/4 + ... where there are n terms 1/2^n followed by n terms -1/2^n. It is easy to group this with each grouped term being zero, but the original series oscillates rather than converging.
A criterion that is adequate for grouping to be valid is that the absolute sum of the terms in each group tends to zero over the sequence of groups. This criterion is true for the valid example given.
Elroch (talk) 22:13, 8 November 2008 (UTC)
replies: text: But your counterexample is not a telescoping series, as the terms don't go positive, negative, positive, negative... If you arrange them that way, you have 1 - 1 + 1/2 - 1/2 + 1/2 - 1/2 + 1/4 - 1/4 + ..., which is truly telescoping and certainly does converge to 0. So there is nothing wrong with the convergence criterion stated in the article, except that it only works for those series covered by the article. -- Jao (talk) 01:12, 17 November 2008 (UTC)

text: The restriction that terms have to be alternately positive and negative does avoid the problem I mentioned. Alternatively, one can have an arbitrary series and group terms together by sign (resulting in a collapsed series whose terms alternate in sign). Elroch (talk) 09:15, 3 December 2008 (UTC)
thread_title:  Pitfalls 

message: I can't find any reliable reference to the term "telescoping product". It would be one of the following:
*  x_i = \frac{y_i}{y_{i+1}} 
:  \prod_{i=m}^n x_i = \frac{y_m}{y_{m+1}} \cdot \frac{y_{m+1}}{y_{m+2}} \cdot \frac{y_{m+2}}{y_{m+3}} \cdot \,\,\cdots\,\, \cdot \frac{y_{n-1}}{y_n} \cdot \frac{y_n}{y_{n+1}} = \frac{y_m}{y_{n+1}}. 
or
*  x_i = \frac{y_i}{y_{i-1}} 
:  \prod_{i=m}^n x_i = \frac{y_m}{y_{m-1}} \cdot \frac{y_{m+1}}{y_m} \cdot \frac{y_{m+2}}{y_{m+1}} \cdot \,\,\cdots\,\, \cdot \frac{y_{n-1}}{y_{n-2}} \cdot \frac{y_n}{y_{n-1}} = \frac{y_n}{y_{m-1}}. 
-- Hugo Spinelli (talk) 00:27, 14 June 2012 (UTC)
thread_title:  Telescoping product 

message: As long as the starting point for the index is finite and the domain of the index is countable (and – for simplivity – non-empty), should not all partial sums have a finite number of terms after cancellation? Moreover, after commutation, regrouping, and some number of evaluations, the number of terms can be reduced from the original N to any positive integer number of terms. I am not sure that this definition is really correct.
I would aim more for "partial sums have a pattern of cancellation between terms" or something. IbexNu (talk) 01:48, 30 January 2021 (UTC)
replies: text: What is more, a partial sum, by definition, only consists of a finite number of terms. So, I'm completely lost reading the first sentence of the article. Madyno (talk) 21:58, 16 March 2021 (UTC)
thread_title:  Definition