# Tag Archives: Zeta

$$\tag{1}1+2+3+4+\cdots =^?-\frac{1}{12}$$ There are many problems with that equality. Can a summation of positive integers lead to a negative quantity ? The equality implies that $$\sum_{k=1}^\infty k = -\frac{1}{12}$$ But we already know that $$S_N = \sum_{k=1}^Nk = \frac{N(N+1)}{2} \to … Continue reading Posted in Zeta function | | Leave a comment ## Relation between Zeta and Dirichlet eta functions proof$$\eta(s) = \left( 1-2^{1-s} \right) \zeta(s) \textit{proof}$$We will start by the RHS$$\left( 1-2^{1-s} \right) \zeta(s) = \zeta(s) – 2^{1-s} \zeta(s)$$Which can be written as sums of series$$\sum_{n=1}^\infty \frac{1}{n^s} – \frac{1}{2^{s-1}}\sum_{n=1}^\infty \frac{1}{n^s}\sum_{n=1}^\infty \frac{1}{n^s} – 2\sum_{n=1}^\infty … Continue reading

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## Integral representation of the zeta function proof

$$\zeta(s) = \frac{1}{\Gamma(s)}\int^\infty_0\frac{t^{s-1}}{e^t-1}dt$$ $$\textit{proof}$$ Start by the integral representation $$\int^\infty_0 \frac{e^{-t}t^{s-1}}{1-e^{-t}}\,dt$$ Using the power expansion $$\frac{1}{1-e^{-t}} = \sum_{n=0}^\infty e^{-nt}$$ Hence we have $$\int^\infty_0\,e^{-t}t^{s-1}\left(\sum_{n=0}^\infty e^{-nt}\right)\,dt$$ By swapping the series and integral $$\sum_{n=0}^\infty\int^\infty_0\,t^{s-1}e^{-(n+1)t}\,dt = \Gamma(s) \sum_{n=0}^\infty \frac{1}{(n+1)^s}=\Gamma(s)\zeta(s)\,$$

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## Relation between polygamma and Hurwitz zeta function proof

$\forall \,\, n\geq 1$ $$\psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z)$$ $$\textit{proof}$$ Use the series representation of the digamma $$\psi_{0}(z) = -\gamma-\frac{1}{z}+ \sum_{n=1}^\infty\frac{z}{n(n+z)}$$ This can be written as the following $$\psi_{0}(z) = -\gamma + \sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+z}$$ By differentiating with respect to … Continue reading

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## Zeta for Even integers proof (Bernoulli numbers)

$$\zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$ $$\textit{proof}$$ We start by the product formula of the sine function $$\frac{\sin(z)}{z} = \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$ Take the logarithm to both sides \log(\sin(z)) – \log(z) = \sum_{n=1}^\infty \log \left(1-\frac{z^2}{n^2 \, \pi^2} … Continue reading

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