# 1 − 1 + 1 − 1 + · · · = 1/2 ??? **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [June 26, 2013, 5:00pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968 "2013-06-26T17:00:39Z")\ **Posts on this page:** 7\ **Page:** 1
**Author:** ![standingwave](https://avatars.discourse-cdn.com/v4/letter/s/9de0a6/32.png) [@standingwave](https://boards.straightdope.com/u/standingwave)\ **Post date:** [June 26, 2013, 5:00pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/1 "2013-06-26T17:00:39Z")
It’s called [Grandi’s series](http://en.wikipedia.org/wiki/Grandi%27s_series) and Numberphile [posted a video](http://youtu.be/PCu_BNNI5x4) on it today. Mind blown and I’m not entirely convinced. First he shows how the infinite series can be shown to be equal to 0. Then he shows how it’s equal to 1. And he finishes up by showing how the best answer is 1/2 and leads into a description of [Thomson’s lamp](http://en.wikipedia.org/wiki/Thomson%27s_lamp). As an engineer, this seemed the most intuitive to me. ---
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [June 26, 2013, 5:09pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/2 "2013-06-26T17:09:20Z")
That series is divergent so there’s no one particular sum which we can say is any better an answer than another one. The 1/2 answer is called the [Cesaro sum](http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation) which, as you note, is just one possible way of doing it. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [June 26, 2013, 5:51pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/3 "2013-06-26T17:51:37Z")
An interesting variant would be if you took the series: 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 … and added parentheses as follows: 1 - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) … or: 1 - 2 - 2 - 2 - 2 - 2 - 2 … so the series works out to negative infinity ---
**Author:** ![standingwave](https://avatars.discourse-cdn.com/v4/letter/s/9de0a6/32.png) [@standingwave](https://boards.straightdope.com/u/standingwave)\ **Post date:** [June 26, 2013, 5:55pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/4 "2013-06-26T17:55:31Z")
> [@friedo](#): > > That series is divergent so there’s no one particular sum which we can say is any better an answer than another one. The 1/2 answer is called the [Cesaro sum](http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation) which, as you note, is just one possible way of doing it. Thanks. So basically, there is no one true answer. It’s been a long time since I studied infinite series (way back in second semester calculus) but I remember that it’s hard to draw definitive conclusions from divergent series. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [June 26, 2013, 6:03pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/5 "2013-06-26T18:03:59Z")
> [@Little\_Nemo](#): > > An interesting variant would be if you took the series: > > 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 … > > and added parentheses as follows: > > 1 - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) … In case it needs to be pointed out, this isn’t in any sense equivalent to the original series, since (by the distributive law) - (1 + 1) would be equivalent to - 1 - 1, not - 1 + 1. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [June 26, 2013, 9:15pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/6 "2013-06-26T21:15:49Z")
As you know, the series 1 + r + r[sup]2[/sup] + r[sup]3[/sup] + … is equal to 1/(1 - r) when |r| \< 1. Through the magic of complex analysis, we can say in a very reasonable sense that that formula holds for any r other than one. If you plug in -1, you get 1/2. The same logic shows that 1 + 2 + 4 + 8 + 16 + … = -1. ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [June 27, 2013, 1:27pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-2/661968/7 "2013-06-27T13:27:40Z")
Since we’re already in Mundane instead of GQ, and since the topic is improper summations, perhaps I can hijack to publish (for the first time ever!) my _“Proof” of the Collatz Conjecture_. 😃 :smack: 😕 Let me stipulate up-front that I’m sure the proof is invalid, though less sure precisely why. The Collatz function operates on odd integers (2x+1) but I’ve represented each such with (x) to work with all integers, rather than just the odds. Thus the sequence    11 → (34 --\>) 17 → (52 → 26 --\>) 13 → (40 → 20 → 10 --\>) 5 → (16 → 8 → 4 → 2) 1 → 1 becomes    5 → 8 → 6 → 2 → 0 → 0 We work with multisets (bundles), with “+” denoting multiset union. B(x) denotes the multiset of integers that map to x, or 4x+1, or 16x+5, or … Because our x=2 corresponds with ordinary 5, B(2) contains all integers that “arrive” at any of 5, 21, 85, … Define B(x) as follows    If x is a positive integer, then      B(x) = {x} + B((2x-1)/3) + B(4x/3) + B(4x+2)    If x is not a positive integer, then      B(x) = null-set We shall prove B(2) = {1,2,3,4,…} = 1 and claim that this is equivalent to the Collatz Conjecture. Consider the B(x) equations, for x \>= 1:   B(1) = {1} + B(6)   B(2) = {2} + B(1) + B(10)   B(3) = {3} + B(4) + B(14)   B(4) = {4} + B(18)   B(5) = {5} + B(3) + B(22)   B(6) = {6} + B(8) + B(26) … Sum all these equations to get    sum (B(x)) = {1,2,3,…} + sum (B(x) except B(2)) It is easily verified that the right side of this sum contains exactly one instance of each B(x) on the left-side _except_ B(2). Subtract “sum B except B(2)” from each side to get    B(2) = {1,2,3,…} Q.E.D.