Talk:Leibniz formula for π

message: The justification of the term-by-term integration here is not actually trivial. Charles Matthews 12:34, 9 Mar 2005 (UTC)
thread_title: Leibniz formula for pi

message: 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} ...
can be rewritten as:
1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} ... - \frac{2}{3} - \frac{2}{7} - \frac{2}{9} ...
ie: as the sum of two overlapping infinite summations, where one is all positive and the other is all negative.
This necessarily converges far worse than the original formulation, as the two sums step at different rates. It works only because the infinities are both of natural numbers and thus equal infinities, so it doesn't matter that the step rate is different. But continuing with it anyway, we get:
\sum_{n=1}^\infty \frac{1}{2n - 1} - \sum_{n=1}^\infty \frac{2}{4n - 1}
Reduce to a single summation:
= \sum_{n=1}^\infty (\frac{1}{2n - 1} - \frac{2}{4n - 1})
Get the denominators equal:
= \sum_{n=1}^\infty (\frac{4n - 1}{(2n - 1)(4n - 1)} - \frac{4n - 2}{(2n - 1)(4n - 1)})
Combine the two fractions:
= \sum_{n=1}^\infty \frac{(4n - 1) - (4n - 2)}{(2n - 1)(4n - 1)}
Simplify:
= \sum_{n=1}^\infty \frac{1}{(2n - 1)(4n - 1)}
Since Pi is equal to four times this, we get that Pi then equals:
4\sum_{n=1}^\infty \frac{1}{(2n - 1)(4n - 1)}
As best as I can tell, this summation is completely useless, but seems easier to scan with the eye than a lot I've seen.  —Preceding unsigned comment added by 75.164.145.226 (talk) 04:07, 17 February 2008 (UTC)
replies: text: Since the series is only conditionally and not absolutely convergent, this whole remark is rubbish from the start. Reordering a conditonally convergent series allows to converge to any real number or to diverge.--LutzL (talk) 19:00, 12 May 2011 (UTC)

text: using same start point and step
\sum_{n=1}^\infty \frac{1}{4n - 3} - \sum_{n=1}^\infty \frac{1}{4n - 1}
simplifies to
\sum_{n=1}^\infty \frac{2}{16n^2-16n + 3} 2600:1003:B84B:D477:14A6:A3D9:895A:5D12 (talk) 18:23, 12 March 2021 (UTC)
thread_title:  Alternative summations 

message: I was trying to find π using a power series in one way or another in order to get the Leibniz formula, and I found this to be a better method of showing how:
We know that \arctan 1 = \frac{\pi}{4}
Take the first derivative of arctan(x) and put it in a geometric series:
\frac{d}{dx} \arctan x = \frac{1}{1 + x^2}
= \sum_{n=1}^\infty (-x^2)^{n-1}
= \sum_{n=1}^\infty (-1)^{n-1} x^{2n-2}
Integrate that:
\int_1^x (-1)^{n-1} t^{2n-2} \, dt = \frac{(-1)^{n-1}}{2n-1} x^{2n-1}
So, we have a power series for arctan(x):
\arctan x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1} x^{2n-1}
Plug in 1 for x and get π/4:
\arctan 1 = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n-1} = \frac{\pi}{4}
For aesthetics (article uses n=0 to ∞):
\arctan 1 = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}
Q.E.D.
-Matt 20:11, 18 March 2006 (UTC)
replies: text: Hint: if N is a power of ten, each term in the right sum will be a finite decimal fraction. I agree that there is room for elaboration in the article. Feel free to edit it. Fredrik Johansson 11:29, 13 September 2006 (UTC)
thread_title:  Better proof 

message: It seems that "Gregory-Leibniz" is the more common ordering.  In particular, "Gregory-Leibniz series" gets
* 27,700 Google hits,
* 41 hits on Google books, and
* 23 hits on Google scholar,
while "Leibniz-Gregory series" only gets
* 9,560 Google hits,
* 8 hits on Google books, and
* 4 hits on Google scholar.
Jim (talk) 16:28, 23 October 2009 (UTC)
thread_title:  Gregory-Leibniz vs. Leibniz-Gregory 

message: "To give the rightful place to this great mathematician, the series should be named 'mAdhava srENi' or Madhava Series. But now Madhava's work is sometimes known as the Madhava–Leibniz series."
Great? Rightful place? Should?  "But now"?  I believe it would be better suited for this articule to discuss other names the series could be known by in the opening paragraph (with redirections as necessary).  History should include all of the history of the series rather than just the history of one "great" mathematician over others.
-Ben0mega (talk) 00:08, 19 May 2010 (UTC)
thread_title:  POV History? 

message: I edited the article to make it shorter and less controversial. I removed the part of the proof that was not a proof. I removed wishful thinking about honoring Madhava. I changed the heading 'history' to 'names'. Bo Jacoby (talk) 06:22, 1 June 2010 (UTC)
thread_title:  an edit 

message: recently boldly moved this article to Gregory's Series and changed the lead paragraph to remove the reference to Leibniz and call it Gregory's formula. "Gregory's series" (note lowercase second word per Wikipedia style conventions) is certainly one of the names of this formula, but "Gregory's formula" or the "Gregory formula" appears to more commonly refer to something else involving numerical integration. In any case I reverted the change feeling that it was the sort of thing we should discuss here first. I have no strong feelings myself on which name we give the article as long as it is one of the common names for this series and is properly capitalized. Any opinions on this besides mine and OliverBel's? There are some discussions immediately above regarding the article title but they are on a different aspect of it (whether to spell out π or use the symbol for it). —David Eppstein (talk) 22:29, 17 May 2014 (UTC)
thread_title:  Article title (again) 

message: The proof, as written, is unacceptably sloppy - large leaps are made without justification, and a new variable (n) is introduced without explanation partway through (leaving the reader to guess as to what it actually is - an arbitrary integer? it is not clear).
Could someone please clean this up?
108.2.128.112 (talk) 03:31, 14 July 2018 (UTC)
thread_title:  Sloppy proof 

message: The series is a sum of rational numbers, which can only result in a rational number.  But π is proven to be an irrational number.
It's apparent that the series therefore cannot EQUAL π (or π/4) since a rational number cannot equal an irrational number. The series must approach π but cannot equal π 
2600:1003:B853:C0F:9512:7625:7F74:ED31 (talk) 02:49, 6 March 2021 (UTC)
replies: text: What makes you think that the sum of an infinite series of rational numbers must be rational? —David Eppstein (talk) 02:56, 6 March 2021 (UTC)

text: It is rational at each finite step, but what makes you think that the result of an infinite process will keep the same properties as its finite steps? In any case this is far off-topic here, because this discussion board is only for discussing improvements to the article based on published reliable sources, not about clearing up your misunderstanding of basic mathematical definitions. —David Eppstein (talk) 04:59, 6 March 2021 (UTC)

text: Series is a limit of partial sums, IT IS NOT A SUM. Just by definition. There is no such concept as infinite sum, at least until summation rules are introduced, like Cesaro summation. 2A00:1FA0:48BD:B662:FD9A:5A04:FE1B:E20D (talk) 12:04, 4 April 2021 (UTC)
thread_title: Series is a rational number