Talk:1 + 1 + 1 + 1 + ⋯

message: 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)
replies: text: Why? Melchoir 09:14, 13 May 2007 (UTC)

text: 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)
thread_title:  Merge 

message: 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 + … = −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)
thread_title: More fun with divergent series

message: 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)
replies: text: 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 ζ(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 ζ(0)=-1/2, because then the function is smooth at that point. —David Eppstein (talk) 02:32, 22 May 2010 (UTC)
thread_title: This doesn't make sense

message: 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)
replies: text: Fixed. Gandalf61 (talk) 07:45, 11 September 2012 (UTC)
thread_title:  Is [[real number]] important? 

message: :
\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)
thread_title: Error?

message: 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)
replies: text: Only if you can find a published reliable source describing and justifying this manipulation. —David Eppstein (talk) 18:15, 21 August 2015 (UTC)
replies: text: 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)

text: 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)
thread_title: Another proof

message: 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)
thread_title:  Another 0=1 problem