Tail of the zeta function \(\zeta(s)\):

\[\displaystyle\sum_{n\geq N}\dfrac{1}{n^s}=\dfrac{N^{1-s}}{s-1}+\dfrac{1}{2N^s}-s\int_N^\infty\left(\{t\}-\dfrac{1}{2}\right)t^{-s-1}\ \mathrm{d}t\]

#ZetaFunction #RiemannZetaFunction #TailOfFunction #Function #Maths #NumberTheory #Analysis

@futurebird
That mathematicians are adding infinite diverging series of positive numbers and coming up with finite negative numbers as the sum, but if I had written that down as an answer in my exams in secondary school, it would have been marked as wrong
https://youtu.be/YuIIjLr6vUA
#YouTube #RiemannZetaFunction #mathematics #maths #math #mathstodon

The Riemann hypothesis, explained by Jørgen Veisdal: 🔗 https://www.cantorsparadise.com/the-riemann-hypothesis-explained-fa01c1f75d3f
\[\zeta(s)=0;\ s\notin2\mathbb{Z}^-\implies\Re(s)=\dfrac{1}{2}\]

#RiemannHypothesis #Riemann #Hypothesis #Atiyah #NumberTheory #Conjecture #AnalyticNumberTheory #Mathematics #CantorsParadise #OpenProblem #UnsolvedProblem #PrimeNumber #RiemannZetaFunction #ZetaFunction #NonTrivialZero #LogarithmicIntegral #XiFunction #PrimeNumberTheorem #PrimeNumberDistribution #PrimeCountingFunction #GammaFunction

PRODUCTS OVER PRIME NUMBERS [2/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\left(1+\dfrac{1}{p(p+1)}\right)=\prod_{p\in\mathbb{P}}\dfrac{1-p^{-3}}{1-p^{-2}}=\dfrac{\zeta(2)}{\zeta(3)}=\dfrac{\pi^2}{6\zeta(3)}\]
\[\displaystyle\prod_{p\in\mathbb{P}}\left(1+\dfrac{1}{p(p-1)}\right)=\prod_{p\in\mathbb{P}}\dfrac{1-p^{-6}}{(1-p^{-2})(1-p^{-3})}=\dfrac{\zeta(2)\zeta(3)}{\zeta(6)}=\dfrac{315}{2\pi^4}\zeta(3)\]
#PrimeProducts #RiemannZetaFunction #EulerProduct #ZetaFunction #InfiniteProduct #NumberTheory

SOME PRODUCTS OVER PRIME NUMBERS [1/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{1}{1-p^{-s}}=\prod_{p\in\mathbb{P}}\left(\sum_{k=0}^\infty\dfrac{1}{p^{ks}}\right)=\sum_{n=1}^\infty\dfrac{1}{n^s}=\zeta(s)\]
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{p^s+1}{p^s-1}=\prod_{p\in\mathbb{P}}\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right)=\sum_{n=1}^\infty\dfrac{2^{\omega(n)}}{n^s}=\dfrac{\zeta(s)^2}{\zeta(2s)}\]
#PrimeProducts #EulerProduct #RiemannZetaFunction #ZetaFunction #InfiniteProduct

SOME PRODUCTS OVER PRIME NUMBERS [1/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{1}{1-p^{-s}}=\prod_{p\in\mathbb{P}}\left(\sum_{k=0}^\infty\dfrac{1}{p^{ks}}\right)=\sum_{n=1}^\infty\dfrac{1}{n^s}=\zeta(s)\]
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{p^s-1}{p^s+1}=\prod_{p\in\mathbb{P}}\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right)=\sum_{n=1}^\infty\dfrac{2^{\omega(n)}}{n^s}=\dfrac{\zeta(s)^2}{\zeta(2s)}\]
#PrimeProducts #EulerProduct #RiemannZetaFunction #ZetaFunction #InfiniteProduct

SOME PRODUCTS OVER PRIME NUMBERS [1/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{1}{1-p^{-s}}=\prod_{p\in\mathbb{P}}\left(\sum_{k=0}^\infty\dfrac{1}{p^{ks}}\right)=\sum_{n=1}^\infty\dfrac{1}{n^s}=\zeta(s)\]
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{p^s+1}{p^s+1}=\prod_{p\in\mathbb{P}}\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right)=\sum_{n=1}^\infty\dfrac{2^{\omega(n)}}{n^s}=\dfrac{\zeta(s)^2}{\zeta(2s)}\]
#PrimeProducts #EulerProduct #RiemannZetaFunction #ZetaFunction #InfiniteProduct

#PrimeProducts #RiemannZetaFunction #InfiniteProduct #AperyConstant