# 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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