## Inequality involving sum of logarithms and hidden zeta-function

I would like to prove the following estimation: if $n \ge 2$ is a natural number, then $$\sum_{k=2}^n \frac{\log^2 k}{k^2} <2 – \frac{\log^2 n}{n}.$$ I have noticed that LHS is indeed bounded by proving that $$\sum_{k = 2}^\infty \frac{\log ^2 k}{k^2} = \zeta”(2) \approx 1.98928$$ and then check with an aid of computer that for … Read more

## Series for ζ(3)−65\zeta(3)-\frac{6}{5}

ζ(2) The inequality 9<π2<10 can be obtained from series ζ(2)=π26=32+12∞∑k=01(k+1)2(k+2)2 and ζ(2)=π26=53−∞∑k=11(k+1)(k+2)2(k+3) Both of them have constant numerator and the denominator a polynomial of fourth degree. ζ(3) Similarly, for Apéry’s constant we can write the inequality 65<ζ(3)<54 from series ζ(3)=65+∞∑k=21k3+4k7 and ζ(3)=54−∞∑k=01(k+1)(k+2)3(k+3) However, the degree of the polynomials is not the same in this case. … Read more

## How does one prove that +∞∑n=11n1+2p≤1+2pp\sum\limits_{n=1}^{+\infty} \frac{1}{n^{1+2p}} \leq \frac{1+2p}{p}?

Got stuck on this while reading a paper. How does one prove that for p>0 the following inequality holds? +∞∑n=11n1+2p≤1+2pp So far I got that the series’ sum is the value of the Riemann ζ-function at z=1+2p. But this gets me no further. Answer Note that since x→1/x1+2p is decreasing in [1,+∞), +∞∑n=11n1+2p=1++∞∑n=21n1+2p≤1++∞∑n=2∫nn−1dxx1+2p=1+∫∞1dxx1+2p=1+12p. AttributionSource : … Read more

## Fourier Transform of the Riemann zeros (Dirac comb)?

Lets assume RH and $\rho_i, i\in\Bbb N$ be the imaginary parts of the non-trivial zeros of the Riemann $\zeta$ function: $\zeta(\frac{1}{2}\pm\imath \rho_i)=0$, $(\forall i)$. Does anonye know if anything (in case what) is known on the (real) Fourier-Transform of a “zeta-zero-Dirac-comb”: $$\mathcal{F}\left \{ \sum_{i=1}^{\infty} \delta(t – \rho_i ) + \delta(t + \rho_i)\right \}[s]$$ … Read more

## Question on convergence of formula for Dirichlet eta function $\eta(s)$

The Dirichlet eta function $\eta(s)$ is related to the Riemann zeta function $\zeta(s)$ as illustrated in (1) below. References (1) and (2) claim formula (2) for $\zeta(s)$ is globally convergent (except where $s=1+\frac{2\,\pi\,i}{\log(2)}n$ and $n\in\mathbb{Z}$) which seems to imply formula (3) for $\eta(s)$ is globally convergent. This is consistent with an answer to one of … Read more

## Prove limn ↦ ∞4nψ(2)(2n)−ψ(2)(1)=2ζ(3)−1\lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1

Prove \lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1\tag{1} where \psi^{(m)}(x) is the polygamma function and \zeta is the Riemann zeta function. This problem was proposed by a friend and solved using the series expansion of digamma function: \psi(x)=\ln x+O\left(\frac1x\right) and by differentiating both sides k times we get \psi^{(k)}(x)=\frac{(-1)^{k-1}(k-1)!}{x^k}+O\left(\frac1{x^{k+1}}\right) and by setting k=2 and x=2^n then letting n\mapsto … Read more

## Infinite product of Zetafunctions

It is well-known (I learned about this first in a video by Papa Flammy) that ∑∞n=2(ζ(n)−1)=1. This result on its own is quite remarkable, but it also implies convergence of the infinite product P=∞∏n=2ζ(n). I somehow got invested in this. Upon trying to work out the exact value of P, I found that P=∞∑n=1c2(n)n, where … Read more

## A new(?) analytic continuation for the Riemann zeta function.

While tweaking the definition for the Euler gamma constant I found that the following appears to be true: ζ(s)=lim when \Re(s)>0, a>0 and b>0. Can you prove it? Answer Use the AFE (approximate functional equation): \zeta(s)=\sum_{k \le x}k^{-s}-\frac{x^{1-s}}{1-s}+O(x^{-\sigma}) uniform in \sigma \ge \sigma_0>0 and valid for say |t| < \pi x, s=\sigma+it Fixing s, \Re … Read more

## Meaning of equality in zeta regularization

It is known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$ but it is also known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$ which can obtained using the analytic continuation of $\zeta(s)$. My question is: What is the true meaning of equality … Read more

## How to prove that Riemann zeta function is zero for negative even numbers? [duplicate]

This question already has answers here: How one can obtain roots at the negative even integers of the Zeta function? (2 answers) Closed 6 years ago. Can anyone please explain to me how to prove that Riemann zeta function is 0 for all negative even numbers. In many references , they have just given the … Read more