Talk:1/2 − 1/4 + 1/8 − 1/16 + ⋯

Untitled

C'mon, this article already has loads of information that would be inappropriate to move to another article. Melchoir 23:25, 16 March 2007 (UTC)[reply]

No, actually, almost all the information is appropriate in some other article. See, for example, Hackenbush, Euler transform, and geometric progression. For example, the Euler transform works on any divergent geometric series, so it should be in that article. — Arthur Rubin | (talk) 01:35, 17 March 2007 (UTC)[reply]

Hackenbush is a general article, and it would have to include information on all infinite games and strings; if one added all that information, there would still have to be an example, and no reason why it shouldn't have its own article. The Euler transform does not "work" on any divergent geometric series; it produces a convergent series only if the original common ratio is in the open disk of radius 2 centered at -1. The fact that the Euler transform happens to take the formal 1/(1+2) to the formal 1/(2+1) is a lucky accident. Yes, this series is a geometric series, but only a geometric series with a common ratio of -1/2 can express the binary expansion of a surreal number.

All of those other articles need to be expanded, but they need generally applicable information, not just the diced up remains of a common example. Melchoir 01:56, 17 March 2007 (UTC)[reply]

I oppose the merge. This series is notable enough to deserve its own entry.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 18:26, 20 March 2007 (UTC)[reply]

I am for this merge, this is a rather small article and this series does not seem to illustrate the kind of importance that would merit its own article--Cronholm144 22:53, 12 May 2007 (UTC)[reply]

Simple Proof

This article was desperately in need of a simple proof like similar articles. Cheers.8.25.32.37 (talk) 11:05, 10 May 2013 (UTC)[reply]

That's not a "simple" proof. — Arthur Rubin (talk) 20:15, 11 May 2013 (UTC)[reply]

Yes it is and moreover it stands alone contrary to other references in the article8.25.32.37 (talk) 20:31, 11 May 2013 (UTC) It is perfectly fine to calculate the value of the series.72.37.134.11 (talk) 14:21, 19 May 2013 (UTC)[reply]

And what you've added is not a "calculation" of the value of the series. What I added is more productive. — Arthur Rubin (talk) 15:12, 19 May 2013 (UTC)[reply]