Talk:Ramanujan summation

message: :2+3+5+7+11+\cdots = 24133 (\Re)
holds as well, but I haven't any article to take citation from. It is P(-1), where P(x) is Prime zeta function (http://mathworld.wolfram.com/PrimeZetaFunction.html).
This is hard to believe. Sums of integers are usually 0 (e.g. 1+4+9+16+...), fractions (e.g. 1+2+3+4+...) or irrational (e.g. 1+2+6+24+...). How did you arrive at that specific integer? I computed a value of 2.92... for the sum of prime numbers. Doubledork (talk) 22:03, 30 July 2009 (UTC)

message: Does anybody know a rationale to how this sum is figured?  Thanks. Root4(one) 05:14, 5 June 2007 (UTC)
thread_title:  Explanation? 

message: Ramanujan did not invent the technique as this article claims. Euler used it long before him.
Look at E55 ("Methodus universalis series summandi ulterius promota") and E352 ("Remarques sur un beau rapport entre les series des puissances tant directes que reciproques ") here http://www.math.dartmouth.edu/~euler/  —Preceding unsigned comment added by 209.67.107.10 (talk) 00:01, 13 June 2009 (UTC)
thread_title:  Not invented by Ramanujan 

message: I found another way to calculate Ramanujan summation:
Suppose we want to calculate f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+\ldots .
Let g\left( x \right) be an analytic continuation of \sum\limits_{m=1}^{x}{\int\limits_{0}^{m}{f\left( t \right)dt}}
Then 
f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+\ldots =-{g}'\left( 0 \right) (\Re).
This method agrees with the regular Ramanujan summation when f\left( x \right) is a polynomial, and this method is also linear, so I suppose it agrees with the regular Ramanujan summation in every series.
77.126.36.120 (talk) 14:32, 30 September 2009 (UTC)
replies: text: Nice, great, but can you deliver proof of this result? If you can so, I'll suggest writing paper rather than forcing solution in talk page of wikipedia article. 82.160.31.58 (talk) 20:56, 8 July 2023 (UTC)
thread_title:  Another way to calculate Ramanujan summation 

message: "For even powers we have:
:1 + 2^{2k} + 3^{2k} + \cdots = 0\ (\Re)"
but 1_−_2_%2B_3_−_4_%2B_·_·_·#Generalizations says:
For even n, this reduces to
:1-2^{2k}+3^{2k}-\cdots = 0.
What's right?
46.115.58.168 (talk) 01:43, 30 August 2012 (UTC)
replies: text: they are BOTH right remember that  \eta(s)= (1-2^{1-s}) \zeta (s)  so they share some zeros. — Preceding unsigned comment added by 85.85.12.24 (talk • contribs) 19:15, 11 September 2013‎
They are both wrong, Ramanujan made a mistake, simple as that. 
Even in the context of Ramanujan Summation, only the 1-1+1- ... = 1/2 and 1-2+3-4 ... = 1/4 are 'correct'. The others do not follow the context and ignore zeroes, and are hence incorrect completely. Ramanujan Summation refers to the impact of the series, and hence zeroes cannot be ignored : http://rahulraj-says.blogspot.in/2014/01/the-power-of-zero.html  — Preceding unsigned comment added by 120.61.22.134 (talk) 17:35, 19 January 2014 (UTC)
thread_title:   + + + +…  or  + - + -…  ? 

message: It doesn't bother anybody else that this article is more or less incomprehensible?  I have no idea what a Ramanujan sum is, how it's calculated, or why it's interesting, and I'm not likely to learn it from the article, since the very first equation has two variables (Bk+1 and Rp) that are not defined anywhere in the article Geoffrey.landis (talk) 21:31, 18 January 2014 (UTC)
thread_title:  Not very comprehensible. 

message: These edits should be reverted; one look at Rahul Raj's cited blog post reveals that he's an amateur (not a bad thing in itself) and doesn't know or understand things beyond Cesaro summation.  — Preceding unsigned comment added by Innerproduct (talk • contribs) 22:07, 19 January 2014 (UTC)
thread_title:  Recent edits citing "Rahul Raj" are not from academic sources 

message: This article needs serious revision Infinite Geometry (talk) 03:58, 17 February 2024 (UTC)
replies: text: I agree. Is there a way to flag this up to the editors? Blitzer99 (talk) 15:33, 22 April 2024 (UTC)
thread_title:  Important note