# Tagged Questions

39 views

### Question about Riemann zeta function + my proof

First let me say that I am 16 years old so I am not very professional in math. English is also a second language so I apologize for any mistakes. Now i have been reading about the Riemann zeta ...
36 views

### A question about the convergence of partial products of zeta of one.

Recently I've been toying around with the Totient function and the Prime Number Theorem and came up with the odd result that the following limit $$\lim_{n\to\infty}\frac{\pi(n)m_n}{\phi(m_n)n}$$ ...
37 views

### Analytic Continuation of the zeta function

Is the analytic continuation of the Riemann zeta function to the upper half plane unique? I don't know much complex analysis, so I can't see why that is the case.
24 views

### Zeta zero sum & reciprocals of prime powers

Below is a plot of $$\dfrac{1}{x}\sum_{n=1}^{75}2\Re\left(\operatorname{Ei}\left(\rho_n\log\left(x\right)\right)\right)$$ where $\rho_n$ is the $n$th zeta zero, with grid lines at primes and prime ...
78 views

### How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$?

How would you show that the Riemann Zeta function, $\zeta(s) < 0$ for $s \in (0,1)$? So far I have that along the critical strip \begin{align} \zeta(s) &= ...
77 views

### Dirichlet series and Riemann zeta function

Im trying to show, for $\Re(s)>1$, that $\displaystyle\sum_{n=0}^{\infty} \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}$, where $d(n)= |\{k \mid k|n \}|$, number of positive integers that ...
41 views

I was looking through the derivation of the Riemann functional equation, and I understand how to obtain the result $$\pi^{-\frac s2} \Gamma (\frac s2) \zeta(s) = \pi^{-\frac{1-s}{2}} ... 1answer 80 views ### Analytic Continuation of Zeta Function using Bernoulli Numbers In my complex analysis textbook by Stein and Shakarchi, as an exercise, I am supposed to extend \zeta(s) to the entire complex plane using Bernoulli numbers, but I am stuck. I can prove that$$ ...
Mathematica knows that the logarithm of $n$ is: $$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$ The von Mangoldt function should then be: ...