Questions on the famed $\zeta(s)$ function of Riemann, and its properties.

learn more… | top users | synonyms

135
votes
11answers
18k views

Why does $1+2+3+\cdots = -\frac{1}{12}$?

$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$. Why should analytically continuing to $\zeta(-1)$ give the right answer?
90
votes
5answers
8k views

What is the Riemann-Zeta function?

In laymen's terms, as much as possible: What is the Riemann-Zeta function, and why does it come up so often with relation to prime numbers?
49
votes
8answers
8k views

Nice proofs of $\zeta(4) = \pi^4/90$?

I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute ...
44
votes
2answers
2k views

Does $\sum _{n=1}^{\infty } \frac{\sin(\text{ln}(n))}{n}$ converge?

Does $\sum _{n=1}^{\infty } \dfrac{\sin(\text{ln}(n))}{n}$ converge? My hypothesis is that it doesn't , but I don't know how to prove it. $ζ(1+i)$ does not converge but it doesn't solve problem here. ...
41
votes
3answers
833 views

How to prove that $\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$

Prove that $$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$ $H_n$ denotes the harmonic numbers.
40
votes
1answer
806 views

A new continued fraction for Apéry's constant, $\zeta(3)$?

As a background, Ramanujan also gave a continued fraction for $\zeta(3)$ as $\zeta(3) = 1+\cfrac{1}{u_1+\cfrac{1^3}{1+\cfrac{1^3}{u_2+\cfrac{2^3}{1+\cfrac{2^3}{u_3 + \ddots}}}}}\tag{1}$ where the ...
37
votes
2answers
1k views

Prime powers, patterns similar to $\lbrace 0,1,0,2,0,1,0,3\ldots \rbrace$ and formulas for $\sigma_k(n)$

Some time ago when decomponsing the natural numbers, $\mathbb{N}$, in prime powes I noticed a pattern in their powers. Taking, for example, the numbers $\lbrace 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 ...
36
votes
0answers
1k views

The log gamma integral $\int_{0}^{z} \log \Gamma (x) \ \mathrm dx$

One way to evaluate $ \displaystyle\int_{0}^{z} \log \Gamma(x) \ \mathrm dx $ is in terms of the Barnes G-function. $$ \int_{0}^{z} \log \Gamma(x) \ \mathrm dx = \frac{z}{2} \log (2 \pi) + ...
33
votes
5answers
2k views

Does $\zeta(3)$ have a connection with $\pi$?

The problem Can be $\zeta(3)$ written as $\alpha\pi^\beta$, where ($\alpha,\beta \in \mathbb{C}$), $\beta \ne 0$ and $\alpha$ doesn't depend of $\pi$ (like $\sqrt2$, for example)? Details Several ...
33
votes
1answer
1k views

Are these zeros equal to the imaginary parts of the Riemann zeta zeros?

Edit 8.8.2013: See this question also. The Fourier cosine transform of an exponential sawtooth wave times $e^{-x/2}$: $$\operatorname{FourierCosineTransform}(\operatorname{SawtoothWave}(e^x)\cdot ...
32
votes
5answers
2k views

What is so interesting about the zeroes of the $\zeta$ function

The Riemann $\zeta$ function plays a significant role in number theory and is defined by $$\zeta(s) = \sum_{n=0}^\infty \frac{1}{n^s} \qquad \text{ for } s > 1 \text{ and } s= \sigma + it$$ The ...
31
votes
2answers
2k views

Books about the Riemann Hypothesis

I hope this question is appropriate for this forum. I am compiling a list of all books about the Riemann Hypothesis and Riemann's Zeta Function. The following are exluded: Books by mathematical ...
30
votes
6answers
1k views

Infinite Series $‎\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$

‎If $f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n$ and $\sum_{n=2}^{\infty}|a_n|$ converges then‎, $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ ‎Since ...
30
votes
1answer
791 views

A Bernoulli number identity and evaluating $\zeta$ at even integers

Sometime back I made a claim here that the proof for $\zeta(4)$ can be extended to all even numbers. I tried doing this but I face a stumbling block. Let me explain the problem in detail here. I was ...
29
votes
2answers
1k views

Proving an “amazing” claim regarding $\zeta( 3)$ and Apéry's proof

I recently printed a paper that asks to prove the "amazing" claim that for all $a_1,a_2,\dots$ $$\sum_{k=1}^\infty\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}=\frac{1}{x}$$ and thus (probably) ...
27
votes
4answers
1k views

Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series.

Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$ Computing the Fourier series of $f$ and ...
26
votes
2answers
747 views

Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

I found the following formula $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$ and it is cited that Euler proved the ...
26
votes
1answer
843 views

Are Primes a Self-Fulfilling Prophecy?

Assume the following process: Let's start with the set of primes $\{p_k\}$ Then we use the Euler product being equivalent to Riemann's Zeta function $$ \prod_{p \text{ prime}} \frac{1}{1-p^{-s}} = ...
24
votes
4answers
1k views

Proving a known zero of the Riemann Zeta has real part exactly 1/2

Much effort has been expended on a famous unsolved problem about the Riemann Zeta function $\zeta(s)$. Not surprisingly, it's called the Riemann hypothesis, which asserts: $$ \zeta(s) = 0 ...
24
votes
2answers
619 views

A double series yielding Riemann's $\zeta$

Can you give me some hints to prove equality: $$\sum_{m,n=1}^{\infty} \frac1{(m^2+n^2)^2} =\zeta (2)\ G-\zeta(4)=\frac{\pi^2}{6}\ G-\frac{\pi^4}{90}$$ where $\zeta (t):= \sum\limits_{n=1}^{+\infty} ...
24
votes
1answer
378 views

Is this integral $\int_0^1\left(\left\{\frac1x\right\}-\frac12\right)\frac{\log(x)}{x}dx$ equal to zero?

My initial question was to find if this integral $$ \int_0^1 \left(\left\{\frac 1x\right\}-\frac12\right)\frac{\log(x)}{x}dx$$ is convergent or divergent. ($\left\{\frac 1x\right\}$ is the fractional ...
23
votes
3answers
1k views

Two Representations of the Prime Counting Function

The bounty for the best work out of Greg's answer, especially the "solving for $\pi^*(x;q,a)$ in terms of all $\Pi^*$ functions (tedious but possible)" part is over. Since Raymond's ...
20
votes
4answers
2k views

Riemann zeta function at odd positive integers

Starting with the famous Basel problem, Euler evaluated the Riemann zeta function for all even positive integers and the result is a compact expression involving Bernoulli numbers. However, the ...
20
votes
1answer
519 views

Upper bound on differences of consecutive zeta zeros

The average gap $\delta_n=|\gamma_{n+1}-\gamma_n|$ between consecutive zeros $(\beta_n+\gamma_n i,\beta_{n+1}+\gamma_{n+1}i)$ of Riemann's zeta function is $\frac{2\pi}{\log\gamma_n}.$ There are many ...
19
votes
2answers
536 views

Formula for $\zeta(3)$ -verification

By simple manipulating with some series I have found the following formula for $\zeta(3)$: $$\zeta(3)=\frac27\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k+2}}{(2k+2)!},$$ where $b_k$ are Bernoulli ...
19
votes
3answers
438 views

Find the value of $\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx$

I'm trying to figure out how to evaluate the following: $$ J=\int_{0}^{\infty}\frac{x^3}{e^x-1}\ln(e^x - 1)\,dx $$ I'm tried considering $I(s) = \int_{0}^{\infty}\frac{x^3}{(e^x-1)^s}\,dx\implies ...
19
votes
5answers
2k views

The main attacks on the Riemann Hypothesis?

Attempts to prove the Riemann Hypothesis So I'm compiling a list of all the attacks and current approaches to Riemann Hypothesis. Can anyone provide me sources (or give their thoughts on possible ...
19
votes
3answers
216 views

Integrals of integer powers of dilogarithm function

I'm interested in evaluating integrals of positive integer powers of the dilogarithm function. I'd like to see the general case tackled if possible, or barring that then as many particular cases as ...
19
votes
2answers
638 views

Riemann's zeta as a continued fraction over prime numbers.

Riemann's zeta function is a function with many faces, I mean representations. I recently derived this one, bellow, as a continued fraction over prime numbers. $$ \zeta(s)=1 ...
18
votes
1answer
353 views

References to integrals of the form $\int_{0}^{1} \left( \frac{1}{\log x}+\frac{1}{1-x} \right)^{m} \, dx$

While extending my calculation techniques, with aid of Mathematica, I found that \begin{align*} \int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{3} \, dx &= -6 \zeta '(-1) ...
17
votes
3answers
580 views

A series involves harmonic number

How do we get a closed form for $$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$
16
votes
2answers
492 views

Simpler zeta zeros

Is it true that $$\lim_{y\rightarrow\infty}\dfrac{\sum_{n=1}^{y}n^{-1/2-iy}}{\zeta(1/2+iy)}=1$$ ? Below is a plot of $$\sum_{n=1}^{y}\dfrac{1}{n^{s}}\text{for }s=\dfrac{1}{2}+iy$$ set against its ...
16
votes
1answer
563 views

“Orientation” of $\zeta$ zeroes on the critical line.

I am pretty ignorant about complex analysis so please forgive my lack of terminology. I saw a pretty picture (posted below) of the behavior of the Riemann zeta function along the critical line. What ...
16
votes
1answer
444 views

An outrageous way to derive a Laurent series: why does this work?

I had to compute a series expansion of $1/(e^{x}-1)$ about $x=0$, and in the course of its derivation, I made a couple of manipulations that are not allowed mathematically. Still, comparing the final ...
16
votes
0answers
219 views

Proof of $\zeta(2)=\frac{\pi^2}{6}$

While messing around with some integrals, I have found the following proof for $\zeta(2)=\frac{\pi^2}{6}$, but I'm not sure if it is valid: We take a look at the integral $I=\int_0^{\frac{\pi}{2}} ...
15
votes
2answers
1k views

Logarithmic derivative of Riemann Zeta function

Given the logarithmic derivative of the zeta function $\dfrac{\zeta^\prime (s)}{\zeta(s)}$ how does it behave near $s=1$? I mean if for $s=1$ the Laurent series for the logarithmic derivative becomes ...
15
votes
1answer
293 views

Infinite Series $\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}$

How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$ I think this post can help me, but I'm not sure.
15
votes
1answer
361 views

A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$

I have done series with $\zeta(2k)$ and $\zeta(k)$, but I have no idea with this one: $$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$$ $\gamma$ is the Euler–Mascheroni Constant. This ...
15
votes
1answer
369 views

What is a zeta function?

In my readings, I've come across a wide variety of objects called zeta functions. For example, the Ihara zeta function, Igusa local zeta function, Hasse-Weil zeta function, etc. My question is simple: ...
15
votes
1answer
393 views

What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions?

We have the following evaluations: $$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = ...
15
votes
3answers
337 views

Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$.

Wolfram$\alpha$ says that we have the following identity $$ \left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}} $$ but, how does one prove such identity?
13
votes
3answers
4k views

Analytic continuation- Easy explanation?

Today, as I was flipping through my copy of Higher Algebra by Barnard and Child, I came across a theorem which said, The series $$ 1+\frac{1}{2^p} +\frac{1}{3^p}+...$$ diverges for $p\leq 1$ and ...
13
votes
4answers
622 views

Do these series converge to logarithms?

It is well known that $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}... =\log(2).$$ If we consider the array: $T(n,k) = -(n-1)\; \text{ if }\; n|k, \;\text{ else } \;1,$ Starting: ...
13
votes
1answer
490 views

challenging integral involving $\zeta(5)$

I ran across a curious integral that seems to be rather tough that some on the site may enjoy. Show that $$\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^{2}}}{1-x^{2}\sin^{2}(x)}dx = ...
13
votes
1answer
331 views

On the zeta sum $\sum_{n=1}^\infty[\zeta(5n)-1]$ and others

For p = 2, we have, $\begin{align}&\sum_{n=1}^\infty[\zeta(pn)-1] = \frac{3}{4}\end{align}$ It seems there is a general form for odd p. For example, for p = 5, define $z_5 = e^{\pi i/5}$. Then, ...
13
votes
2answers
857 views

How to show that the Laurent series of the Riemann Zeta function has $\gamma$ as its constant term?

I mean the Laurent series at $s=1$. I want to do it by proving $\displaystyle \int_0^\infty \frac{2t}{(t^2+1)(e^{\pi t}+1)} dt = \ln 2 - \gamma$, based on the integral formula given in Wikipedia. ...
13
votes
1answer
727 views

Derivative of Riemann zeta, is this inequality true?

Is the following inequality true? $$\gamma -\frac{\zeta ''(-2\;n)}{2 \zeta '(-2\;n)} > \log (n)-\gamma$$ This for $n$ a positive integer, $n=1,2,3,4,5,...$, and more precisely when $n$ approaches ...
13
votes
1answer
386 views

Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$

Recently, I was able to prove that $$\sum_{n=1}^\infty \frac{1}{n \binom{2n}{n}}= \frac{\pi}{3\sqrt{3}}$$ $$\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}}= \frac{\pi^2}{18}$$ But does anybody know ...
13
votes
3answers
3k views

Factorial of infinity

So, I've read in this article that: $$\zeta'(0) = \log\sqrt\frac{1}{2\pi}$$ And that: $$e^{-\zeta'(0)} = 1\cdot2\cdot3\cdot\ldots\cdot\infty = \infty! = \sqrt{2\pi}$$ I found this result very ...
12
votes
1answer
896 views

What exactly *is* the Riemann zeta function? [duplicate]

I'm doing a little project on the $\zeta$ function, and I am at a complete loss of what it is actually doing. I understand it is way over my head, but when I am plugging say $\zeta(1 + i)$ into ...