Talk:Geometric series

On the example at the start where the factor is 2/3 it says "Consider the sum of the following geometric series:". Is the wording wrong? Can you have a sum of a series? or should it say "Consider the sum of the following sequence". Or since they are putting s = 1 + 2/3+ ... should it just say "Consider the following series"? I am not advanced enough in math to be confident in changing the text.

-Adam Siembida

The "sum of the series" is the standard word used in this case. It requires a definition, that you will find in the article Series (mathematics). --Bdmy (talk) 09:24, 15 May 2009 (UTC)

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Why the factor a everywhere?? --85.5.93.40 (talk) 22:28, 7 March 2011 (UTC)

This is how series are often presented in Calculus texts. While students should just be taught to factor out "garbage" like factors of a, I don't think a rewrite of the article is useful. Austinmohr (talk) 07:49, 25 May 2011 (UTC)

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Your article conflicts with geometric progression

Ooops ! I'm wrong this is really a serie.

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Maybe you mean "a series"? "Series", with a final "s", is either singular or plural in English; there is no word "serie". -- Mike Hardy

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Thanks I did'nt know. "Une série", "des séries" I believe it was the same in English. :)

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I copied over the piece on geometric series in geometric sequence. It still needs to be integrated into the article. On further thought I feel that Geometric sequence and Geometric series should be combined since they deal with each other. Therefore I'm putting them together under Geometric progression and redirecting.

-Mathematical error- The series is given as a sum from k = 0 to infinity of r^k for |r|<1 but when r is zero the first term would be 0^0 which is undefined. it should be 1 + a sum from k = 1 to infinity of r^k for |r|<1

-Please try to make this understandable to non experts The use of strange symbols with no definition is a drag for those of us who are not educated in math-language. Please give a definition of that greek symbol (is it sigma? or something else) - and not in mathese - but in regular language. Since this an encyclopedia and not a technical journal please keep the hermetic language to a minimum. Thankyou Canuckistani 22:43, 5 January 2009 (UTC) —Preceding

unsigned comment added by Canuckistani (talk • contribs)

Some links for references.

After being momentarily confused by one of the images I thought I might improve it. Here are some links to "geometric" methods for finding the sum of a geometric series.

• Proof without Words: Geometric Sums by Warren Page. http://www.jstor.org/stable/2689632

• J. H. Webb, $\sum \limits _{n=0}^{\infty }ar^{n}={\frac {a}{1-r}}$, Proof without words, http://www.jstor.org/stable/2689568

• Proof without Words by Benjamin G. Klein and Irl C. Bivens. http://www.jstor.org/stable/2689356

• Proof without Words: Geometric Series by Elizabeth M. Markham http://www.jstor.org/stable/2690738 (I like one of the presentations here better then ours)

• Proof without Words: Geometric Series by Sunday A. Ajose and Roger B. Nelsen http://www.jstor.org/stable/2690617 (This contains almost exactly one of our images, should probably reference this.)

• Proof without Words: An Alternating Series by James O. Chilaka http://www.jstor.org/stable/2691280 (This one is very interesting because it is the only to address an r<0 that I have found so far.)

• Proof without Words: Geometric Series by The Viewpoints 2000 Group http://www.jstor.org/stable/2691106

That's all for now.

proof:

I think the proof should be changed to reflect that s=1+r+..r^(n-1) versus rs=r+r^2+r^3...+r^n hence s = (r^n-1)/(r-1) —Preceding unsigned comment added by 192.18.192.76 (talk) 10:26, 6 May 2009 (UTC)

As the first line of this article says, it is about infinite sums. -- Jitse Niesen (talk) 10:47, 6 May 2009 (UTC)

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Does anybody have some information about the double geometric series, given by $\sum \sum _{m\geq 1,n\geq 1}x^{mn}=\sum _{m\geq 1}{\frac {x^{m}}{1-x^{m}}}$, with |x|<1? In particular, what happens if x is close to 1 Are there asymptotic formula available in that case? 140.112.50.253 (talk) 03:06, 8 May 2009 (UTC)

I believe that when x goes to 1, with 0 < x < 1 one has that the sum is equivalent to

${\frac {1}{1-x}}\ln {\Bigl (}{\frac {1}{1-x}}{\Bigr )},$

by comparing to Riemann sums for ${\frac {1}{1-t}}$ on the interval $[0,x],$ with division points $x,x^{2},x^{3},\ldots$ --Bdmy (talk) 13:10, 8 May 2009 (UTC)

To be more precise, I think that as x tends to 1, one has

${\Bigl (}\sum _{m\geq 1}{\frac {x^{m}}{1-x^{m}}}{\Bigr )}-{\frac {1}{1-x}}\ln {\Bigl (}{\frac {1}{1-x}}{\Bigr )}\sim {\frac {c}{1-x}}$

for some positive constant $c\sim 0.57\dots$ that I am not able to express in closed form. --Bdmy (talk) 13:30, 9 May 2009 (UTC)

Actually I am now sure that $c$ is equal to the Euler-Mascheroni constant, and that the expansion above must be written somewhere, and extended to more terms. --Bdmy (talk) 14:47, 9 May 2009 (UTC)

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on Zeno's paradox:

The hidden assumption is that a sum of infinite number of finite steps can not be finite. This is of course not true as evident by the convergence of the geometrical series with r=1/2 illustrated at the picture at the introduction section of this article.

I don't think this is right. That a finite distance can be divided into infinitely many steps is not a hidden assumption, it is the main premise of the argument. The paradox (to show why this is absurd) is that it would take an infinite amount of time to complete these steps - here is where the real "hidden" assumption comes in - that each step requires a finite and bounded amount of time and therefore the whole process would take forever. Zeno's paradox is resolved if it is assumed that the time to perform each step can be made infinitesimally small, allowing convergence.

N. —Preceding unsigned comment added by 70.49.87.164 (talk) 04:45, 1 July 2009 (UTC)

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Formula in section 2.2

The formula in section 2.2 says it is "the sum of the first n terms of a geometric series". It actually is the sum of the first n+1 terms of a geometric series.

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—Preceding unsigned comment added by 88.81.99.141 (talk) 17:17, 9 April 2010 (UTC)

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Introductory Series Expression and Series Illustration Do Not Agree

Please consider either modifying the 1/2^n series expression or the 1/4^n series illustration so both represent the same series or specify the 1/4^n series in the illustration's caption. Persons using this page to acquaint themselves with the subject will naturally associate the expression and the illustration. The fact that 1/2 appears as the first term in the expression and as a dimension in the illustration could be especially confounding. Thank You. VoilàY'all (talk) 23:52, 25 June 2010 (UTC)

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NOTE: I subsequently added information to the illustration caption to help distinguish series expression in the text from the series summation in the illustration. VoilàY'all (talk) 00:26, 26 June 2010 (UTC)

Archimedes' quadrature of the parabola proof

Actually there is error in image. If triangle area is 1, then scale is not 1 under triangle, but another. With Heron's formula it apears that if a=b=c=1, then area $S={\sqrt {{1+1+1 \over 2}\cdot (1.5-1)\cdot (1.5-1)\cdot (1.5-1)}}={\sqrt {1.5\cdot 0.5^{3}}}={\sqrt {0.1875}}=0.433012701.$

The area enclosed by a parabola and a line is the union of infinitely many triangles.

If area is 1, then there is quick way to find a=b=c values by looking how much times is different area 1 from area 0.433, so difference is 1/0.433012701=2.309401077. So normaly if triangle have a=1 and h=1, then his area S=0.5*1*1=0.5, and if triangle have a=4 and h=4, then triangle are is 16 times bigger, S=0.5*4*4=8. Or if triangle a=2 and h=2, then S=0.5*2*2=2, so 4 times bigger. Thus if area is 2.3 times bigger, then $a={\sqrt {2.309401077}}=1.519671371$ times bigger, so this is exact a=b=c=1*1.519671371=1.51967 value. (for more approving this thinking let's say a=2, h=4, s=0.5*2*4=4 and if [lines are 2 times longer] A=4, H=8, S=0.5*4*8=16, value is 4 times bigger; or if a=4, h=2, s=0.5*4*2=4, by multiplying A=a*3=4*3=12, H=h*3=2*3=6, S=4*9=0.5*12*6=36, so 36/4=9 times, 3*3=9, get it?)

So to proof this series

$1\,+\,2\left({\frac {1}{8}}\right)\,+\,4\left({\frac {1}{8}}\right)^{2}\,+\,8\left({\frac {1}{8}}\right)^{3}\,+\,\cdots .$

need to find h, which with pythagor theorem is easy, because I know now, that a=b=c=1.519671371. So half of a is d=a/2=1.519671371/2=0.759835685. So by Phytagor theorem $h={\sqrt {b^{2}-d^{2}}}={\sqrt {1.51967^{2}-0.7598^{2}}}={\sqrt {1.732050807}}=1.316074013.$ So what kind of parabola in image is? It is $y=-x^{2}+h=-x^{2}+1.316074013$ parabola. Values x are [-0.759835685; 0.759835685] and values y are [0; 1.316]. By integrating I should get area S>1, so:

$S=\int _{-0.7598}^{0.7598}(-x^{2}+1.316074013)dx=(-{x^{3} \over 3}+1.316074013\cdot x)|_{-0.7598}^{0.7598}=$

$=(-{(-0.7598)^{3} \over 3}+1.316074013\cdot (-0.7598))-(-{0.7598^{3} \over 3}+1.316074013\cdot 0.7598)=$

$=(0.146230445-1)-(-0.146230445+1)=1.707539109$.

Also I want to try another integration way by finding area under parabola "lines" and then substracting it from area s=a*h=1.519671371*1.316074013=2. So parabola then is $y=x^{2}$. And so integration is simplier:

$S=\int _{-0.7598}^{0.7598}x^{2}dx={x^{3} \over 3}|_{-0.7598}^{0.7598}={(-0.7598)^{3} \over 3}-{(0.7598)^{3} \over 3}=-0.292460891.$

So P=s-|S|=2-0.292460891=1.707539109, this is the area under the parabola like in image.

Sides a, b and c may not be equal length (just looks so in wrong image). But why in article no information about What parabola it is? Maybe it's y=ax*x+c, maybe y=x*x+c, maybe y=x*x+bx+c, hm?

Anyway, Let's see what area under parabola in image will be with those series:

$P=1+2\cdot {1 \over 8}+4\cdot {1 \over 8^{2}}+8\cdot {1 \over 8^{3}}+16\cdot {1 \over 8^{4}}+32\cdot {1 \over 8^{5}}+64\cdot {1 \over 8^{6}}+128\cdot {1 \over 8^{7}}=$

$=1+0.25+0.0625+0.015625+0.00390625+0.000976562+0.00024414+0.000061035=1.333312988.$

So if I flip parabola in image and put it highest point onto point (x; y)=(0; 0), then I will have parabola y=x*x. So then I just will find y highest point with puting into x farthest point $y=0.759835685^{2}=0.577350268$ (because seems this series should be correct even if for triangle if a=b=c). So now h=0.577350268. And triangle area should be 1. So S=a*h=1.519671371*0.577350268=0.875823827. Somthing wrong, it apears that triangle is too small with area s=a*h/2=0.437911913.

I think found where is problem. In article there no any information what kind of parabola it is. If it is y=ax*x+bx+c then all calculations becomes wrong. Just why in article no information for what kind parabola you will get, you kinda make parabola around any triangle, calculate parabola area but don't know what kind this parabola is, this is problem.

Let's try to proof in another way. Let's take parabola, which I choosing, say I TAKING PARABOLA y=x*x and x==[-1; 1] and y==[0; 1]. So triangle inside parabola have base a=2 and h=1. Triangle area is s=a*h/2=2*1/2=1. Now with integration method I will find area under parabola lines:

$S=\int _{-1}^{1}x^{2}dx={x^{3} \over 3}|_{-1}^{1}={-1 \over 3}-{1 \over 3}=-{2 \over 3}=-0.66666$. So Rectangle area is S=a*h=2*1=2. And so parabola area is [rectangle area minus area under parabola lines] P=a*h-|-0.6666|=2-0.6666=1.333333.

Now by using series I will try to get same answer and it seems that I previously got it (1.333312988) and triangle area is equal to 1. So those series are now proved. And it means, that those series only good for parabola y=x*x.

Now I just want to see, is it correct for parabola y=x*x, when x==[-3; 3] and y==[0; 9]. But then first need to find triangle area of this kinda big parabola. So a=6 and h=9, so triangle area is s=a*h/2=6*9/2=27, so triangle area is 27 times bigger than in previous case. So Parabola area should be P=1.33333*27=35.99999=36. So let's check it by integrating:

$U=\int _{-3}^{3}x^{2}dx={x^{3} \over 3}|_{-3}^{3}={(-3)^{3} \over 3}-{3^{3} \over 3}=-9-9=-18.$

So here now I got area |U|=18 under parabola lines. Rectangle area is twice bigger than triangle R=a*h=6*9=54. And so parabola area from point (x; y)=(0; 0) to highest point [also [to] triangle base] is P=R-|U|=a*h-|U|=54-18=36. So just perfect, answer the same. Geometrical series for parabola area finding completely proved.

Geometric View

I removed the section below. It is unclear whether the author is addressing a mistake in the figure as it appears here or as it appears in the cited text. Moreover, how the figure relates to the geometric sum is not self-explanatory, and so should be elaborated upon if re-included.

Here is a geometric way of looking at the geometric series from E.Hairer and G.Wanner, Analysis by Its History, section III.2, FIGURE 2.1, page 188, Springer 1996 - there is a mistake in the figure: it should read r/(1-r), then add the missing term 1 to get 1/(1-r) :

For completeness, when $r=1$, the sum of the first n terms is:

$a+a1^{1}+a1^{2}+a1^{3}+\cdots +a1^{n-1}=\sum _{k=0}^{n-1}a=an.$

Austinmohr (talk) 07:44, 25 May 2011 (UTC)

The "economics" section

By my reckoning the output at the bottom of the "economics" section should be 100*(1+I)/I, as opposed to 100/I as it currently stands. — Preceding unsigned comment added by 67.189.78.224 (talk) 00:54, 16 October 2011 (UTC)

Merger proposal

I propose that Geometric progression be merged in part or whole into Geometric series. The concept of the progression or sequence is necessary to understand the series, so it is necessary in the Series article. For some reason, the series article concerns infinite series, but finite series are described in the geometric progression article. This is confusing because it makes it sound as though Geometric series refers to the infinite progression, and never a sum of a finite sequence. Thelema418 (talk) 02:28, 24 August 2012 (UTC)

Althought a geometric progression is very different to a geometric series (one is a sequences, the other is a series), most of the geometric progression article seems to talk about geometric series. Provided Thelema418 is prepared to conduct the merger that s/he has proposed, I have no reservations about a merger taking place. — Fly by Night (talk) 22:19, 25 August 2012 (UTC)

Let us not to repeat the mistake with square (algebra). The proposed direction of a merger is egregiously incorrect. "Progression" is algebra, one can understand progression without infinity, and progressions are really used without any convergence, e.g. as polynomial rings. "Series" is a calculus and, in some sense, topology. These mathematical structures are very different. Of course, the current "geometric progression" has a lot of off-topic, but this should not encourage users to make articles progressively worse, as was made a year ago with redirecting square (algebra) to square number. Incnis Mrsi (talk) 09:13, 10 September 2012 (UTC)

I agree. Geometric progressions and geometric series belong to different parts of mathematics, and should be kept separate for roughly the same reason that the Addition and Series (mathematics) articles are separate. In particular, this article is part of Category:Calculus, while the Geometric progression article is not.Jim.belk (talk) 04:56, 26 January 2014 (UTC)

I would agree; they should be together as are Arithmetic Progression and Arithmetic Series. It's confusing to have Geometric Progression and Geometric Series as separate, lengthy articles. Startswithj (talk) 18:48, 30 November 2013 (UTC)

I don't think the analogy with Arithmetic Progression and Arithmetic Series is relevant. This article is primarily about infinite geometric series, which is the most common use of the term "geometric series" in mathematics. Indeed, most mathematicians would refer to the sum of a finite geometric progression as a "geometric sum" or a "finite geometric series". There is no concept of an "infinite arithmetic series", which is why there's no corresponding article on Arithmetic Series. Jim.belk (talk) 04:56, 26 January 2014 (UTC)