Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [summation-method]

Use for methods for constructing generalized sums of series, generalized limits of sequences, and values of improper integrals.

In mathematical analysis, the need arises to generalize the concept of the sum of a series (limit of a sequence, value of an integral) to include the case where the series (sequence, integral) diverges in the ordinary sense. This generalization usually takes the form of a rule or operation, and is called a summation method.

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Towards a new proof of infinitude of primes ( with possible unified application to other primes of special forms whose Infinitude is unknown):

I'm trying to analyse the primes with the following point of view Consider the following partial sum : $$S(p)=\sum_{n=2}^p\sin^2\left(\frac{π\Gamma(n)}{2n}\right)$$ The summand is zero for non-primes greater than 5 , and finite and non-decreasing…
bambi
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On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$

Ramanujan found the following formula: $$\large \sum_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$$ I let $e^{2\pi n}-1=\left(e^{\pi n}+1\right)\left(e^{\pi n}-1\right)$ to try partial fraction decomposition and turn the sum into…
Mr Pie
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$\sum_{n=1}^\infty\frac{n}{(2n-1)16^n}\binom{2n}{n}^2\left(\sum_{k=n}^\infty\frac{2^k}{k\binom{2k}{k}}\right)=1-\sqrt2+\log(1+\sqrt2).$

Prove:$$\sum_{n=1}^\infty\frac{n}{(2n-1)16^n}\binom{2n}{n}^2\left(\sum_{k=n}^\infty\frac{2^k}{k\binom{2k}{k}}\right)=1-\sqrt2+\log(1+\sqrt2).$$ I'm sorry that I don't even know how to start. I haven't met this kind of series before. I've learnt…
user765381
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Can we show that $1+2+3+\dotsb=-\frac{1}{12}$ using only stability or linearity, not both, and without regularizing or specifying a summation method?

Regarding the proof by Tony Padilla and Ed Copeland that $1+2+3+\dotsb=-\frac{1}{12}$ popularized by a recent Numberphile video, most seem to agree that the proof is incorrect, or at least, is not sufficiently rigorous. Can the proof be repaired to…
ziggurism
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Approximation of a summation by an integral

I am going to approximate $\sum_{i=0}^{n-1}(\frac{n}{n-i})^{\frac{1}{\beta -1}}$ by $\int_{0}^{n-1}(\frac{n}{n-x})^{\frac{1}{\beta -1}}dx$, such that $n$ is sufficiently large. Is the above approximation true? If the above approximation is true, by…
Hasan Heydari
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Sum of reciprocal sine function $\sum\limits_{k=1}^{n-1} \frac{1}{\sin(\frac{k\pi}{n})}=?$

The question comes to me when I find there are answers on summation of some forms of trigonometric functions, i.e. $$ \sum\limits_{k=1}^{n-1} \frac{1}{\sin^2(\frac{k\pi}{n})}\\ \sum\limits_{k=0}^{n-1} \tan(\frac{k\pi}{n})\\ $$ Sum of the reciprocal…
Ethanabc
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Is there an identity for $\sum_{k=0}^{n-1}\csc(w+ k \frac{\pi}{n})\csc(x+ k \frac{\pi}{n})\csc(y+ k \frac{\pi}{n})\csc(z+ k \frac{\pi}{n})$?

What I'd like to find is an identity for $$\sum_{k=0}^{n-1}\csc\left(w+ k \frac{\pi}{n}\right)\csc\left(x+ k \frac{\pi}{n}\right)\csc\left(y+ k \frac{\pi}{n}\right)\csc\left(z+ k \frac{\pi}{n}\right)$$ Further I wonder if a method can be extended…
onepound
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Is there an identity for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$?

Is there a simple relation for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$ like there is for $\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)$? Looking at Jolley, Summation of Series, formula…
onepound
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Summation methods ordered by strength

A summation method is a partial function from scalar sequences to scalars, i.e. an element of the set $\mathbb{C}^\mathbb{N} \rightharpoonup \mathbb{C}$. A summation method $\Sigma_1$ is weaker than a summation method $\Sigma_2$ iff $\Sigma_1…
user76284
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Exercise 2 from Terry Tao's blog on Euler-Maclaurin, Bernouilli numbers, and the zeta function

In the blog post The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic continuation, Terry Tao looks at the commonly-cranked 'absurd' formulae $$\begin{align} \sum_{n \geq 1} 1 &= -1/2 \tag{1} \\ \sum_{n \geq…
BigThumb
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Integral form(s) of a general tetration/power tower integral solution: $\sum\limits_{n=0}^\infty \frac{(pn+q)^{rn+s}Γ(An+B,Cn+D)}{Γ(an+b,cn+d)}$

In many tetration/power tower integrals, one sees a general form of the following. Let this new function be notation used to show the connection between the general result and special cases using types of Incomplete Gamma functions. The goal is to…
Тyma Gaidash
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A Summability methods which sum the harmonic series

Studying summability theory I've come across many summation methods however by now I know only two not very interesting method which re-sums the harmonic series $\sum_{n=0}^\infty \frac 1{n+1}$ : the null method (most trivial of all, sums every…
AlienRem
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A method for evaluating sums/discrete functions by assuming they can be made continuous and differentiable?

Suppose I had a function that satisfied the property $f(x)=f(x-1)+g(x)$. For any $x\in\mathbb N$, it is easy enough to see that this boils down to the statement $$f(x)=f(0)+\sum_{k=1}^xg(k)$$ If we return to our functional equation and…
Simply Beautiful Art
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Is there an identity for $\sum_{k=0}^{n-1}\csc\left(x+ k \frac{\pi}{n}\right)\csc\left(y+ k \frac{\pi}{n}\right)$?

What I'd like to find is an identity for $$\sum_{k=0}^{n-1}\csc\left(x+ k \frac{\pi}{n}\right)\csc\left(y+ k \frac{\pi}{n}\right)$$ here it can be shown that where $x=y$, $$n^2 \csc^2(nx) = \sum_{k=0}^{n-1}\csc^2\left(x+ k \frac{\pi}{n}\right)$$ I…
onepound
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Do we have to use Bernoulli polynomials in the Euler-Maclaurin summation formula?

It may be that I have not picked up the proof, but I cannot see where the third condition of Bernoulli polynomials, given below, is used in the derivation of the Euler-Maclaurin summation formula. The Bernoulli polynomials are defined inductively…
Sputnik
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