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Questions tagged [ramanujan-summation]

Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to infinite divergent series.

Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to infinite divergent series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined. It is like a bridge between summation and integration.

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Does Ramanujan summation evaluate the series $\sum \frac{1}{n^s}$ to $\zeta(s)$ or $\zeta(s)-\frac{1}{s-1}$?

On Wikipedia, in the article on Ramanujan summation as well as some related articles, examples of Ramanujan summation of the form $ \sum\frac{1}{n^s}$ are done for various values of $s$ which seem to imply that Ramanujan summation yields…
ziggurism
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A Ramanujan-like summation: is it correct? Is it extensible?

I'm still exercising with summation-procedures which I try to make correct Ramanujan-summations. I'm looking at the (gap-)series $$ s(1/2,2) = (1/2)^1+(1/2)^{4}+(1/2)^{9}+(1/2)^{16}+(1/2)^{25}+... $$ and more general at $$ s(b,p) =…
Gottfried Helms
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The most complex formula for the golden ratio $\varphi$ that I have ever seen. How was it achieved?

I am fascinated by the following formula for the golden ratio $\varphi$: $$\Large\varphi = \frac{\sqrt{5}}{1 + \left(5^{3/4}\left(\frac{\sqrt{5} - 1}{2}\right)^{5/2} - 1\right)^{1/5}} - \frac{1}{e^{2\pi\,/\sqrt{5}}}\,\mathop{\LARGE…
Mr Pie
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A dig at Ramanujan's: $\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0$

Ramanujan's claim on page 98, in the book (`Ramanujan's note book part 1' by Bruce C. Berndt) is that $$\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,\qquad p>0\tag{1}\label{theclaim}$$ The proposed proofs for $p=1,2$…
Z Ahmed
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Proving $\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))$ for an identity of Srinivasa Ramanujan

Ramanujan is supposed to have given more than five thousand elegant results, a good number of them are yet to be proved or disproved. Yesterday in the comment section of Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac…
Z Ahmed
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What is the Ramanujan summation for the series $\sqrt[n]{2}$

A Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series In my case, I'm interested in assigning a value to the divergent series $$\sum_{n=1}^\infty f(n) \ \ \ \ \ \…
Graviton
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Mellin transform of theta function $\theta$ to show $\zeta(-1)=-\frac{1}{12}$

Edit: I realize a lot of attention is pointed at the proclamation $\zeta(-1)=-\frac{1}{12}$. Im more interested in the derivation of the Zeta function using the methods describe below, thank you! Before you read, I'm requesting a hint as to how to…
Batmannilsson
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Why/How was Hardy impressed with Ramanujan’s work prior to seeing any proofs of their validity?

Hardy’s famous correspondence with Ramanujan highlights an interesting aspect of the mathematics discipline; the appreciation of a mathematical expression prior to knowing its validity. Why/how was Hardy impressed with Ramanujan’s early work on…
Cybernetic
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Did Gosper or the Borweins first prove Ramanujans formula?

In 1985, Gosper used the not-yet-proven formula by Ramanujan $$\frac{ 1 }{\pi } = \frac{2\sqrt{2}}{99^2}\cdot \sum_{n=0}^\infty \frac{(4n)!}{(n!)^4}\cdot\frac{26390 n+1103}{99^{4n}}$$ to compute $17\cdot10^6$ digits of $\pi$, at that time a new…
L. Milla
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Why does $1+2+3+4+\dots=-\frac1{12}$ in a couple different ways?

$1+2+3+4+\dots$ is undefined when using regular summation. If you use either Ramanujan summation or Zeta function regularization, then $1+2+3+4+\dots=-\frac1{12}$. This article lists some over definitions of summation, and they all give the same…
Christopher King
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Closed form for $\sum_{n=1}^\infty \frac{1}{n(e^{2 n \pi}-1)}$

I need to calculate the sum $$\sum_{n=1}^\infty \frac{1}{n(e^{2 n \pi}-1)}$$ Write $$S=\sum_{n=1}^\infty \frac{1}{n(e^{2 n \pi}-1)}$$ $$S=\sum_{n=1}^\infty\frac{1}{n} \sum_{m=1}^\infty e^{-2 \pi n m}$$ Now changing the order of summation…
Max
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How does one get that $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$?

While watching interesting mathematics videos, I found one of the papers of Srinivasa Ramanujan to G.H.Hardy in which he had written $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$. The problem is that every term on the left is more than $\frac{1}{120}$ yet…
Vidyanshu Mishra
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Interpretation of Ramanujan summation of infinite divergent series

I am not mathematician by any means so this question might be rather stupid. I came across this Wikipedia article on Ramanujan's summation and found this bewildering formula, $$1 + 2 + 3 + \dots = - \frac1{12}$$ The article also says that…
hashbrown
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Sum of the divisors of $n$, related to the Hardy-Littlewood circle method

Prove that $$\sigma(n) = \frac{\pi^2}{6}\,n\sum_{q = 1}^\infty q^{-2}c_q(n)$$ where $$ c_q(n) = \sum_{a = 1, (a, q) = 1}^q \exp(2\pi i an/q)$$ and $\sigma(n)$ is the sum of the divisors of $n$. Also, how does this relate to the Hardy-Littlewood…
Mayank Pandey
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number coefficients of an infinite root of 2

Today I stumbled upon one of Ramanujan's infinite roots with all the integers and that got me curious so I started to try to create one of my own. I started with $2=2$. Then $$2=\sqrt4$$ Then $$2=\sqrt{1+3}$$ Then $$2=\sqrt{1+\sqrt{9}}$$ Then…
Brothersquid
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