\infty}\left[\sum_{k=2}^N \Lambda(k)
\ln\left({{k+1} \over k}\right)-\ln N \right].   \eqno(120)$$

On the right side of Eq. (119), the dominant asymptotic order $\ln^{j+1}
N/(j+1)$ is cancelled by the sum, leading to a new set of constants analogous to the second and third terms on the right side of Eq. (7), whose explicit determination remains open. Again partial summation may be applied to the right side of Eq. (119) and alternative forms written. As we do not wish to invoke the Riemann hypothesis, we do not pursue this here.

An approximate value of the first contribution on the right side of Eq. (120) is $0.2419488514703058$, enabling a numerical approximation for the constant coming from the limit term. In comparison, the term $\zeta'(2)/[2\zeta(2)]=(\gamma-\ln 2+\ln \pi+12\ln A)/2
\simeq -0.2849804965472664$ where $A$ is Glaisher’s constant.

For the proof of Proposition 6 we form $${1 \over {j!}}\sum_{m=2}^\infty {{(-1)^m} \over m}\left({{\zeta'} \over \zeta}\right)^{(j)}(m)=-{{(-1)^j} \over {j!}}\sum_{k=2}^\infty \Lambda(k)\ln^j 
k\left[{1 \over k} -\ln\left({{k+1} \over k}\right)
\right].  \eqno(121)$$ We then add the expression (118) and Eq. (119) follows.

The importance of a new representation of the $\eta_j$ coefficients lies especially with new representation and estimation of the critical alternating binomial sum $$S_2(n) \equiv -\sum_{m=1}^n {n \choose m} \eta_{m-1}
=\sum_{m=1}^n (-1)^m {n \choose m} 