Talk:1/2 + 1/4 + 1/8 + 1/16 + ⋯

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Other articles with similar titles (for proof of precedent):

What's really silly about the way this page stands at the moment is the way it says at the top "This article needs additional citations for verification." —Preceding unsigned comment added by 87.240.229.43 (talk) 10:25, 12 April 2011 (UTC)


Questions:

  • Why does this deserve its own entry, rather than a mention under a more general article on infinite series?
  • Who in their right mind would go searching for an article under this name?

Thanks. --Finngall 19:21, 15 March 2007 (UTC)

    • Why would anyone be "searching for any article by name"? Your supposed rhetorical question is stupid and it's implied criticism irrelevant. I just got here by googling "1/2+1/4+1/8" , the fact that this is the title here presumably helped get it to the top of the G results. — Preceding unsigned comment added by 95.176.83.179 (talk) 09:53, 29 May 2011 (UTC)


Answers:

  • This series is notable because it is a representation of one of the famous paradoxes of Zeno of Elea.
  • Not relevant. This is the standard title of series without common names. --ĶĩřβȳŤįɱéØ 19:34, 15 March 2007 (UTC)

For the record, I agree with the proposed deletion. This article is about a straightforward convergent infinite geometric series, which does not deserve an article in its own right - at best it could be an example in geometric series. But if there is going to be a debate about this, let's take it to AfD - that's what the process is for. Gandalf61 10:07, 16 March 2007 (UTC)

Support for merge to Geometric Progression, maybe a sub section can be made there for special cases? Leigao84 17:33, 8 May 2007 (UTC)

The Zeno's paradox connection isn't the only unique application of this series; it is also the famous 0.111… in binary, and it receives specific attention in education research. Granted, the current article doesn't hint at these things, but it could. Melchoir 18:55, 16 March 2007 (UTC)

  • \frac{1}{\frac{1}{a^1}+\frac{1}{a^2}+\frac{1}{a^3}+\frac{1}{a^4}+\frac{1}{a^5}\ldots}=\frac{1}{\sum_{n=1}^{\infty}\frac{1}{a^n}}=a-1;a>0

user:twentythreethousand22:00, 27 September 2007

Did you have something to say? Melchoir 01:14, 28 September 2007 (UTC)
  • This is just a proof of how to build a rational numbers or infinite numbers.

\frac{1}{7}\times{7}=1,or{0.142857}\times{7}=999999

user:twentythreethousand3:15, 29 September 2007

The Series[edit]

Alright, so the notation of the series used here is

 \sum_{k=0}^\infty \frac12\left(\frac12\right)^k = \frac12\frac{1}{1-\textstyle\frac12} = 1.

However, I intuitively think of the series as 1 + 1/2 + 1/4 + 1/8 + ... = 2 - starting at 1 is cleaner, just as in other series, for example 1 + 1/2 + 1/3 + 1/4 + ... But if we started the series as 1/2 + 1/4 + 1/8 + ..., then the notation should be

 \sum_{k=1}^\infty \left(\frac12\right)^k = \frac{\frac12}{1-\textstyle\frac12} = 1.

Majopius (talk) 02:38, 16 April 2010 (UTC)

I understand your feelings, but the point here is the principle of least surprise (for the reader wishing to learn, rather than the editor writing the article). Please compare the notation here with the one at Geometric_series#Formula. Regards, Paradoctor (talk) 16:18, 16 April 2010 (UTC)
Perhaps it would be best to generalize the series there as well i.e.  \sum_{k=N}^\infty x^k = \frac{x^N}{1-x}? Cshanholtzer (talk) 18:29, 2 September 2010 (UTC)


Wouldn't it make sense to actually display the result in the article? Maybe even at the top? I know it's just a *2 multiplication away, but still it seems strange that the result appears nowhere explicitly.
 \sum_{k=0}^\infty (\frac12)^k = 2.

Regards, Paedric (talk) 16:41, 29 October 2012 (UTC)

Distraction graphic must go[edit]

Please remove the extravagant bouncing graphic. It is a serious hindrance to actually reading the article. Neither is is presence or content particularly enlightening. A static graphic if you must, but even that is not really that useful.

Far too many animated images now cluttering WP. — Preceding unsigned comment added by 95.176.83.179 (talk) 09:56, 29 May 2011 (UTC)

I disagree. It's referred to in the text; I would prefer a black and white graphic showing cumulative subtotals in the square, but I consider the graphic helpful. — Arthur Rubin (talk) 14:36, 29 May 2011 (UTC)

Awful proof[edit]

The proof provided in this article is awful. The obvious elegant proof is a follows: Let X = 1/2 + 1/4 + 1/8 + 1/16 ... Then 2X = 1 + 1/2 + 1/4 + 1/8 + 1/16... = 1 + X Therefore X = 1

I will let someone good at math symbols insert this in the text. Cheers 8.25.32.37 (talk) 21:40, 9 May 2013 (UTC) Just added the elegant proof in the text8.25.32.37 (talk) 01:21, 10 May 2013 (UTC)

That's not particularly helpful, either. — Arthur Rubin (talk) 20:14, 11 May 2013 (UTC)

Your personal sentiment does not matter, this article lacks a direct stand alone proof — Preceding unsigned comment added by 8.25.32.37 (talk) 01:46, 13 May 2013 (UTC) This is perfectly fine to calculate the value of the series.72.37.134.11 (talk) 14:20, 19 May 2013 (UTC)

This article has an accurate stand-alone proof, which your addition is not. — Arthur Rubin (talk) 15:13, 19 May 2013 (UTC)
That doesn't mean there's anything wrong with a simple proof. Wikipedia is an encyclopedia for the general reader, so simple, accessible explanations are appropriate, regardless of whether they're mathematically rigorous. NE Ent 16:13, 19 May 2013 (UTC)
  1. It's not a proof; at best it's a demonstration that if the sequence converges, it has the specified value.
  2. It's difficult to read. The combination of inline text and mathematical expressions is jarring.
  3. It repeats the correct proof in the section above. If it is to be included, then the discussion of geometric series in the "lead" should be eliminated. I guess I'll go that way with it, if you insist on adding the "simple proof". — Arthur Rubin (talk) 05:52, 20 May 2013 (UTC)
I've reformatted the "proof" so as to be readable, although it's still not completely correct, and removed the duplication of geometric series#Sum. I still think a less-trimmed version of the original lead, without the duplication of "simple proof", is preferable, but NE Ent (talk · contribs) has given a non-specious reason for inclusion, even though not accurate. — Arthur Rubin (talk) 06:04, 20 May 2013 (UTC)

Glad that common sense finally prevails8.25.32.37 (talk) 03:29, 25 May 2013 (UTC)

It does not equal 1[edit]

This proof is flawed. At first someone could easily notice that it is obviously not equal to 1. The "simple proof" posted seems to make sense but it doesn't. One problem there may be the way that infinity is defined. It is very misleading to regular people, but since as far as math goes, using number 1 as the answer is very good for performing calculations with it.

1/2 + 1/4 + 1/8 + 1/16 ..... 1/2^(∞-2) + 1/2^(∞-1) + 1/2^∞ ( 1/4 + 1/8 + 1/16 + 1/32 ..... 1/2^(∞-2) + 1/2^(∞-1) + 1/2^∞ )*2 = 1/2 + 1/4 + 1/8 + 1/16 ..... 1/2^(∞-3) + 1/2^(∞-2) + 1/2^(∞-1) And not the same as the top one.

Then you can notice that due to the fact of how "infinity" ( ∞ ) is defined in math that it is the same but really it isn't.

∞    1      ∞      1
∑  ----- =  ∑  ---------

X=1 2^X X=1 2^(∞-X+1)

∞      1
∑  --------- = UNDEFINED

X=1 2^(∞-X+1)

∞    1
∑  ----- = UNDEFINED, NOT 1

X=1 2^X

^- To view this properly click to edit this section, the math equations will show up properly.

Grumble. We've dealt with this before. See, for example .999.... — Arthur Rubin (talk) 12:19, 14 August 2013 (UTC)
I am starting to believe that it isn't worth it discussing this with others. It obviously is not 1. This really makes me sad, seeing how someone would believe that it really is 1. 0.999... is not equal to one. 1>0.999... which is really the truth. The proof that attempts to prove it uses a faulty assumption, you can not multiply infinity. 0.999.... cannot be multiplied because it never ends. But multiplication by 10 is also believed to be simply moving the decimal point. That isn't what multiplication is, especially not with imaginary numbers. — Preceding unsigned comment added by 24.191.67.15 (talk) 03:35, 15 August 2013 (UTC)
It may not be worth discussing it, because your statement is obviously false in "conventional" mathematics. — Arthur Rubin (talk) 22:32, 15 August 2013 (UTC)