Talk:Telescoping series

Change Title to Telescoping Sums?

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)

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)[reply]

Gosper

My recollection is that Bill Gosper introduced the idea of telescoping. He is one of the pioneers of computer symbolic mathematics programs, having contributed to both Macsyma and Mathematica, for example. His web page [1] cites A calculus of Series Rearrangements, which might be the place to look. --KSmrq^T 01:52, 3 December 2005 (UTC)[reply]

What!!!??? How could he have introduced this idea if his life was so recent that he worked with electronic computers?? Michael Hardy 22:28, 3 December 2005 (UTC)[reply]

OK, it is asserted at User talk:KSmrq that the term has been found in a 1957 document on mathscinet. Someone is telling me that it's in a math dictionary published in 1949. More to follow when I get more information. My guess is the term originated between 1850 and 1910 -- but that's just a guess. Telescoping sums were used by Euler in the 18th century, but I suspect not by that name. Michael Hardy 22:46, 5 December 2005 (UTC)[reply]

And to be clear, I completely agree that series were being rearranged long before Gosper. One of the challenges of formalization was to understand what rearrangements were safe, as pursued by Karl Weierstrass and others. The questions for this article are, what distinguishes "telescoping", when and where did the term originate, and how is it used today. With respect to current use, Gosper and others have made significant contributions to algorithms used in modern symbolic computer algebra systems, especially with regard to hypergeometric series. In this context the term "creative telescoping" is used. [2] A helpful paper is

Abramov et al. "Telescoping in the context of symbolic summation in Maple". Journal of Symbolic Computation, v38 (2004), 1303–1326. (PDF)

The extraordinary generality of hypergeometric series and the power of these modern algorithms [3] [4] [5] allows the discovery of new and valuable identities. See

Petkovsek, Wilf, Zeilburger. A=B. (Book available online. [6])

for further details. I apologize if my previous brief remark confused or annoyed the historians! --KSmrq^T 22:08, 6 December 2005 (UTC)[reply]

Why use essentially the same example twice?

"Example 1" is essentially the same as the earlier example. Michael Hardy 22:33, 3 December 2005 (UTC)[reply]

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)[reply]

Rewrite

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)[reply]

Proof?

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)[reply]

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)[reply]

What the final value is, is certainly stated explicitly in the article. Michael Hardy 21:55, 5 September 2007 (UTC)[reply]

Applications?

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).[reply]

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)[reply]

Pitfalls

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)[reply]

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)[reply]

I would take "telescoping" to mean that in finite partial sums every term cancels except the first and the last, or the last two, or some bounded number of terms. ("Bounded" means never exceeding some number that does not grow as the number of terms being summed grows.) The article isn't very precise about that. I'll think about how best to phrase a definition. Michael Hardy (talk) 02:26, 17 November 2008 (UTC)[reply]

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)[reply]

So far there is NOTHING wrong with this statement: 0 = ∑0 = ∑(1 -1) = ∑(-1 +1)

But where did you get 1? ... = 1 +∑(-1 +1) =1?

Maybe there are indeed divergent examples of telescoping series. But this one is just wrong.

In my own opinion, the argument for a 'pitfall' here seems forced. — Preceding unsigned comment added by 180.194.238.164 (talk) 04:59, 17 December 2013 (UTC)[reply]

I'm removing it as it doesn't make any sense whatsoever. Someone's inability to do maths is not a pitfall of telescoping series. — Preceding unsigned comment added by Smidderwibh (talk • contribs) 14:26, 5 September 2016 (UTC)[reply]

Telescoping product

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)[reply]

Incorrect?

From the pitfall section: "is not correct because this regrouping of terms is invalid unless the individual terms converge to 0;" The alternating harmonic series has individual terms which converge to zero, but regrouping of terms is not permissable there. Shouldn't this say that the partial sum of the individal terms must be absolutely convergent?

Definition

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)[reply]

Oops, sometimes, it cannot be reduced to more than 2 terms. And, I suppose, it could be reduced to 0 terms, if all of them cancel somehow (easy example: $\sum _{n}(n-n)$ ). IbexNu (talk) 01:51, 30 January 2021 (UTC)[reply]

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)[reply]

I have changed the opening paragraph. The salient feature of a telescoping series is that the general term is presented in a way such that consecutive terms cancel---again, it is about the *presentation* of the series, not the series itself. Any series *can* be rewritten as a telescoping sum. — Preceding unsigned comment added by 206.207.16.4 (talk) 22:18, 12 April 2023 (UTC)[reply]

Who wrote the example section of the article?

Like what

It's full of grammar mistakes and talks about nonsense. 147.83.201.99 (talk) 07:36, 30 October 2024 (UTC)[reply]

Thanks for for editing in better english and adding xls table as example checking the formula many mathematicians seems you must don't be informed was already discovered many years ago... StefanoMaruelli (talk) 06:35, 31 October 2024 (UTC)[reply]

I totally agree. The entire Example section is a grammatical nightmare. It is also not an example so much as a diatribe. I started to make corrections until I realized the entire thing is just a mess. I do not have time to rewrite it now but perhaps someone can get started on it? LRDodd (talk) 08:55, 12 November 2024 (UTC)[reply]