Conjecture: The Euler-Mascheroni constant (γ) is the limit of the real part of the Riemann...

Discussion in 'Mathematics' started by Cran Cowan, Oct 8, 2018.

1. Cran CowanGuest

Does this make sense? (sorry for the formatting. I'm kind of new at LaTex)

Proof that: $\gamma = \lim_{t \rightarrow 0} Re \big\{\zeta(1 + it)\big\}$ ..................................(1)

Given Eq. 22 from http://mathworld.wolfram.com/RiemannZetaFunction.html

$\zeta (s) = \frac{1}{s-1} \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k} (k+1)^{1-s}$ ..............(2)

Substitute: $s = 1 + it$ ...............................................................(3)

$\zeta (1 + it) = \frac{1}{it} \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k} (k+1)^{-it}$ .........(4)

Expanding the exponential:

$\zeta (1 + it) = \frac{-i}{t} \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k} \big(\cos(-t \ln (k+1)) + i \sin(-t \ln (k+1))\big)$ ....(5)

$\zeta (1 + it) = \frac{1}{t} \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k} \big(\sin(-t \ln (k+1)) - i \cos(-t \ln (k+1))\big)$ ......(6)

Taking the real part of (6):

$Re \big\{\zeta (1 + it)\big\} = \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k}\frac{ \sin(-t \ln (k+1))}{t}$ ................................(7)

Taking the limit for both sides of (7) as t approaches 0:

$\lim_{t \rightarrow 0} Re \big\{\zeta (1 + it)\big\} = \lim_{t \rightarrow 0} \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k} \frac{ \sin(-t \ln (k+1)) }{t}$ .......(8)

By L'Hôpital's rule: $\lim_{t \rightarrow 0} \frac{ \sin(-t \ln (k+1))}{t} = -\ln (k+1)$ .........................................(9)

Hence, $\lim_{t \rightarrow 0} Re \big\{\zeta (1 + it)\big\} = \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k} {n\choose k}(- \ln (k+1) )$ .....(10)

Manipulating indices k and n...

$\lim_{t \rightarrow 0} Re \big\{\zeta (1 + it)\big\} = \sum_{n=0}^{\infty} \frac{1}{n+1} \sum_{k=0}^{n} (-1)^{k+1} {n\choose k} \ln (k+1)$ ........(11)

$\lim_{t \rightarrow 0} Re \big\{\zeta (1 + it)\big\} = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{k=0}^{n-1} (-1)^{k+1} {n-1\choose k} \ln (k+1)$ .......(12)

From Eq. 56 at http://mathworld.wolfram.com/Euler-MascheroniConstant.html

$\gamma = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{k=0}^{n-1} (-1)^{k+1} {n-1\choose k} \ln (k+1)$ ............................................(13)

Subtracting (12) from (13) yields (1):

$\gamma = \lim_{t \rightarrow 0} Re \big\{\zeta(1 + it)\big\}$ .......................................................................QED

Cran Cowan 6 Oct 2018

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