Talk:1 + 1 + 1 + 1 + ⋯

Fun with divergent series
This is just for fun, Melchoir. If

1 + 1 + 1 + 1 + … = ½, and

1 &minus; 1 + 1 &minus; 1 + … = ½, then by adding them we obtain

2 + 2 + 2 + 2 + … = 1, and by subtraction we obtain

2 + 2 + 2 + 2 + … = 0; and from the two of these we obtain the very useful equation

0 = 1.

I remember running across a "demonstration" that 0 = 1 in a little book that warned against the dangers of division by zero when I was just a kid. It's too bad that book didn't mention divergent series! ;^> DavidCBryant 22:24, 8 March 2007 (UTC)


 * Nice! There's a similar thing you can do to get negative one half in the first place:
 * $$x = 1+1+1+1+\cdots = 0+1+0+1+\cdots$$
 * $$-2x = 0-2+0-2+\cdots$$
 * $$-x = x + -2x = 1-1+1-1+\cdots$$
 * $$-x = 0+1-1+1-\cdots$$
 * $$-2x = 1+0+0+0+\cdots = 1$$
 * $$x = -\frac12$$
 * Actually, this nonproof has something to do with the relation between the zeta and eta functions.
 * But for the record, I'm sure that zeta function regularization is horribly non-linear and breaks every axiom that both of us appear to have just used! Melchoir 22:36, 8 March 2007 (UTC)

Problem with the first "proof that 0=1". You haven't justified the addition and subtraction of an infinite series, as well as your reordering of the terms.-- Ķĩřβȳ ♥  ♥  ♥  Ťįɱé  Ø  23:16, 13 April 2007 (UTC)


 * I know that, Kirby. (I didn't exactly re-order the terms, as in scramble them up ... I just omitted a lot of zeroes.) Anyway, this is just for fun. The whole thing about divergent series is sort of "just for fun". That is, one can derive useful results from divergent series, but one must be very careful when doing so. For instance, there are divergent series for things like the logarithmic integral that actually converge for the first ten or fifteen terms before they begin to diverge, and give rather close approximations to the right values when truncated exactly at the point where they start to diverge. DavidCBryant 20:10, 14 April 2007 (UTC)


 * Yes, omitting zeroes is the problem. Addition actually gives us
 * 2 + 0 + 2 + 0 + 2 + 0 + &hellip; = 1, while subtraction yields
 * 0 + 2 + 0 + 2 + 0 + 2 + &hellip; = 0.
 * To equate them you need stability. HTH.  –Dan Hoey 21:50, 15 April 2007 (UTC)

Merge
This article should be merged with the zeta function. Then made a redirect. Disamibig is also a possibility--Cronholm144 22:31, 12 May 2007 (UTC)
 * Why? Melchoir 09:14, 13 May 2007 (UTC)

It is a weak stub, why not? not to be trite but I have addressed this elsewhere.--Cronholm144 09:19, 13 May 2007 (UTC)
 * In this case, I can give you a very explicit example of why not: the article Zeta function doesn't need an anecdote on the viewpoints of physicists, as expressed during seminars in Barcelona, on one particular "value" of the formal Dirichlet series expansion... of the zeta function. It's too far off topic. Melchoir 09:54, 13 May 2007 (UTC)

Let's move to your talk page--Cronholm144 09:57, 13 May 2007 (UTC)

More fun with divergent series
It just occurred to me that this article solves the St. Petersburg paradox. For me to participate in the game, you have to pay me a quarter:

½ + ½ + ½ + ½ + … = &minus;¼

But hey Melchoir, I am willing to make you an offer you can't refuse: I'll repetitively subject myself to the game for free... ;) JocK 14:20, 28 July 2007 (UTC)

This doesn't make sense
Here's the problem:

x = 1 + 1 + 1 + 1 + ... x = 1 + 1 + (1 + 1 + 1 + 1 + ...) x = 1 + 1 + x 0 = 2

This proves that any number equals any other number, which, by common logic, doesn't make any sense. -Sixeightyseventyone (talk) 01:37, 22 May 2010 (UTC)
 * Yes, you can do this with any divergent series. Different summation methods will come up with different answers. Nevertheless, it's sometimes useful to come up with a value for this one. For instance, what is the value &zeta;(0) of the Riemann zeta function when its argument zero? You can argue as above that it's undefined but it turns out to be more convenient to set &zeta;(0)=-1/2, because then the function is smooth at that point. —David Eppstein (talk) 02:32, 22 May 2010 (UTC)

This problem relates to treating infinity as if it were just another number, i.e., assuming that the algebraic properties of real numbers (associative, commutative, etc...) also hold for ∞. Replace x with infinity, ∞, in your argument above, and it's easier to locate where things go wrong:

∞ = 1 + 1 + 1 + 1 + ... ∞ = 1 + 1 + (1 + 1 + 1 + 1 + ...) ∞ = 1 + 1 + ∞ 0 = 2

For the last step, one must assume that $x &minus; x$ = 0, or equivalently, $∞ &minus; ∞$ = 0, but this is not necessarily the case. $∞ &minus; ∞$ simply cannot be defined. Consider the infinite set of positive integers, $ℤ$, and then subtracting the infinite set of even numbers, E; the result, $ℤ \ E$, is an infinite set, i.e., the "odd" numbers: |ℤ| &minus; |E| = $∞ &minus; ∞$ = ∞. So, is this the only possible answer? No! Consider the infinite set of positive integers, $ℤ$, and then subtracting the infinite set of numbers greater then 2, T ; the result $ℤ \ T$ = {1, 2}, so: |ℤ| &minus; |T| = $∞ &minus; ∞$ = 2. And likewise, one can arrive at any integer (or infinity) as being the result of $∞ &minus; ∞$. This is related to other Paradoxes of infinity. Justin W Smith talk/stalk 03:54, 22 May 2010 (UTC)

Is real number important?
Although I do not object against fixing my poor English, the disappearance of the "real number" link after [] is not good. It is the crucial point in understanding why series which diverge for 19th-century mathematics may converge in modern mathematics (especially, from topological point of view). Incnis Mrsi (talk) 14:02, 10 September 2012 (UTC)


 * Fixed. Gandalf61 (talk) 07:45, 11 September 2012 (UTC)

Error?


\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) $$ I don't understeand much of mathemathics, but sin(pi*s/2) isn't equal to 0 when s = 0? That should make the whole product equal to 0. (Excuse me for my english).


 * Yes, but the simple zero of sin(z) at z=0 is cancelled by the simple pole of zeta(z) at z=1, so the whole product is nonzero. Melchoir (talk) 17:53, 27 March 2015 (UTC)

It seems like pure nonsense, maybe it's some sort inside joke. Shouldn't be on wikipedia without a huge disclaimer. The expansion of Zeta(1-s) as (1/s + ...) is completely ignoring the given formula for the expansion. Given that the sin-sum converges against 0, and the given Zeta(1-s)-sum converges against the alternating harmonic series, which itself converges to ln(2), anything but zero for the product needs a much better explanation.

Another proof
Perhaps this could be added into the article?

To get 1 + 1 + 1 + 1 + ... = -1/2

First consider Grandi's series: 1 - 1 + 1 - 1 + ... = 1/2

Subtract Grandi's series from our series:

(1 + 1 + 1 + ...) - (1 - 1 + 1 - ...) = 0 + 2 + 0 + 2 + ... (1 + 1 + 1 + ...) - (1/2) = 2 * (1 + 1 + 1 + ...) => (1 + 1 + 1 + ...) = -1/2

Genieieiop (talk) 18:11, 21 August 2015 (UTC)
 * Only if you can find a published reliable source describing and justifying this manipulation. —David Eppstein (talk) 18:15, 21 August 2015 (UTC)
 * With luck, it ought to be possible to find a source that carries out this manipulation, more or less, to compute the zeta regularized sum. Melchoir (talk) 19:44, 21 August 2015 (UTC)

Thinking more about it, I think this manipulation might be incorrect. The sum of Grandi's Series and 1 + 1 + 1 + ... also yields 2 + 0 + 2 + 0 + ..., twice of 1 + 1 + 1 +... . It seems like the zeroes in between each term can contribute to a difference in the sum. Any thoughts on this? Genieieiop (talk) 15:51, 1 September 2015 (UTC)


 * Adding zeroes to a divergent series almost always changes its regularized sum, even for the simplest series and summability methods. For example, by diluting Grandi's series, it's easy to shift its Cesaro sum to 1/3 or 2/3. You can think of this as analogous to rearranging the terms of a conditionally convergent series. The Riemann rearrangement theorem doesn't prevent us from manipulating conditionally convergent series; it just tells us that we have to be careful not to apply inappropriate manipulations. Melchoir (talk) 19:00, 1 September 2015 (UTC)

Another 0=1 problem
Let's take two sums: S1 = 1 + 2 + 3 + 4 + 5 + ... = -1/12 S2 = 1 + 1 + 1 + 1 + 1 + ... = -1/2 S1 - S2 = 1 + 2 + 3 + 4 + 5 + ... - 1 - 1 - 1 - 1 - 1 - ... = 0 + 1 + 2 + 3 + 4 + ... = 0 + S1. So S1 - S2 = 0 + S1. If we agree that S1 = -1/12 and S2 = -1/2, than -1/12 - 1/2 = 0 - 1/12 -1/2 = 0 And then by dividing everything by -1/2 we get 1 = 0. What could be done to solve the problem? --Maks-s15 (talk) 18:53, 1 February 2016 (UTC)