Natural log of #2 turns up frequently. It's sort of cattycorner to pi, although related.
\(
\begin{array}{|c|c|c|c|c|c|c|}
\hline & \frac{\zeta(1)}{1} & \frac{\zeta(2)}{2}& \frac{\zeta(3)}{3}& \frac{\zeta(4)}{4} & \frac{\zeta(5)}{5}&\dots\\
\hline \lim\limits_{x\to 1^+} log(\frac{x}{x-1}) & \frac{1}{1 \cdot 1^1} & \frac{1}{2 \cdot 1^2} & \frac{1}{3 \cdot 1^3}& \frac{1}{4 \cdot 1^4}& \frac{1}{5 \cdot 1^5}&\dots\\
\hline log (\frac{2}{1}) & \frac{1}{1 \cdot 2^1} & \frac{1}{2 \cdot 2^2} & \frac{1}{3 \cdot 2^3}& \frac{1}{4 \cdot 2^4}& \frac{1}{5 \cdot 2^5}&\dots\\
\hline log (\frac{3}{2}) & \frac{1}{1 \cdot 3^1} & \frac{1}{2 \cdot 3 ^2} & \frac{1}{3 \cdot 3^3}& \frac{1}{4 \cdot 3^4}& \frac{1}{5 \cdot 3^5} &\dots\\
\hline log (\frac{4}{3}) & \frac{1}{1 \cdot 4^1} & \frac{1}{2 \cdot 4^2} & \frac{1}{3 \cdot 4^3}& \frac{1}{4 \cdot 4^4}& \frac{1}{5 \cdot 4^5} &\dots\\
\hline log (\frac{5}{4}) & \frac{1}{1 \cdot 5^1} & \frac{1}{2 \cdot 5^2} & \frac{1}{3 \cdot 5^3}& \frac{1}{4 \cdot 5^4}& \frac{1}{5 \cdot 5^5} &\dots\\
\hline \dots &\dots & \dots&\dots &\dots&\dots&\dots\\
\hline
\end{array}
\)
Log shift function, with x ? N a ? R : [0,1]:
\(log (\frac{x}{x-1}) \to log(\frac{x+1}{x}) \,= \, f(x,a) =\, \sum\limits_{n=1}^\infty \frac{1}{n\cdot x^n} - \frac{a}{n \cdot x^{2n}}\)
Shift logs by a=1:
\(
\begin{array}{|c|c|c|c|c|c|c|}
\hline & \frac{\zeta(1)}{1} & -\frac{\zeta(2)}{2}& \frac{\zeta(3)}{3}& -\frac{\zeta(4)}{4} & \frac{\zeta(5)}{5}&\dots\\
\hline log(\frac{2}{1}) & \frac{1}{1 \cdot 1^1} & -\frac{1}{2 \cdot 1^2} & \frac{1}{3 \cdot 1^3}& -\frac{1}{4 \cdot 1^4}& \frac{1}{5 \cdot 1^5}&\dots\\
\hline log (\frac{3}{2}) & \frac{1}{1 \cdot 2^1} & -\frac{1}{2 \cdot 2^2} & \frac{1}{3 \cdot 2^3}& -\frac{1}{4 \cdot 2^4}& \frac{1}{5 \cdot 2^5}&\dots\\
\hline log (\frac{4}{3}) & \frac{1}{1 \cdot 3^1} &- \frac{1}{2 \cdot 3 ^2} & \frac{1}{3 \cdot 3^3}& -\frac{1}{4 \cdot 3^4}& \frac{1}{5 \cdot 3^5} &\dots\\
\hline log (\frac{5}{4}) & \frac{1}{1 \cdot 4^1} & -\frac{1}{2 \cdot 4^2} & \frac{1}{3 \cdot 4^3}& -\frac{1}{4 \cdot 4^4}& \frac{1}{5 \cdot 4^5} &\dots\\
\hline log (\frac{6}{5}) & \frac{1}{1 \cdot 5^1} & -\frac{1}{2 \cdot 5^2} & \frac{1}{3 \cdot 5^3}& -\frac{1}{4 \cdot 5^4}& \frac{1}{5 \cdot 5^5} &\dots\\
\hline \dots &\dots & \dots&\dots &\dots&\dots&\dots\\
\hline
\end{array}
\)
zeta to eta shift function, s ?N, a ? R: [0,1]:
\( \frac{\zeta(s)}{s} \to \frac{\eta(s)}{s} \, = g(s,a) = \, \sum\limits_{n=1}^\infty \frac{1}{s \cdot n ^ s} - \frac{a}{s \cdot (2n)^ s}\)
\(
\begin{array}{|c|c|c|c|c|c|c|}
\hline & \frac{\eta(1)}{1} & -\frac{\eta(2)}{2}& \frac{\eta(3)}{3}& -\frac{\eta(4)}{4} & \frac{\eta(5)}{5}&\dots\\
\hline log(\frac{2}{1}) & \frac{1}{1 \cdot 1^1} & -\frac{1}{2 \cdot 1^2} & \frac{1}{3 \cdot 1^3}& -\frac{1}{4 \cdot 1^4}& \frac{1}{5 \cdot 1^5}&\dots\\
\hline - log (\frac{3}{2}) & -\frac{1}{1 \cdot 2^1} & \frac{1}{2 \cdot 2^2} &- \frac{1}{3 \cdot 2^3}& \frac{1}{4 \cdot 2^4}& -\frac{1}{5 \cdot 2^5}&\dots\\
\hline log (\frac{4}{3}) & \frac{1}{1 \cdot 3^1} &- \frac{1}{2 \cdot 3 ^2} & \frac{1}{3 \cdot 3^3}& -\frac{1}{4 \cdot 3^4}& \frac{1}{5 \cdot 3^5} &\dots\\
\hline -log (\frac{5}{4}) &- \frac{1}{1 \cdot 4^1} & \frac{1}{2 \cdot 4^2} &- \frac{1}{3 \cdot 4^3}& \frac{1}{4 \cdot 4^4}& -\frac{1}{5 \cdot 4^5} &\dots\\
\hline log (\frac{6}{5}) & \frac{1}{1 \cdot 5^1} & -\frac{1}{2 \cdot 5^2} & \frac{1}{3 \cdot 5^3}& -\frac{1}{4 \cdot 5^4}& \frac{1}{5 \cdot 5^5} &\dots\\
\hline \dots &\dots & \dots&\dots &\dots&\dots&\dots\\
\hline
\end{array}
\)
\(log(\frac{2}{1}) -log (\frac{3}{2}) + log (\frac{4}{3}) -log (\frac{5}{4}) +... \, = \,log(\frac{2}{1}) + log (\frac{2}{3}) + log (\frac{4}{3}) +log (\frac{4}{5}) +log (\frac{6}{5}) +... \)
From the above, from the Wallis product for pi:
\( \sum\limits_{n=1}^\infty \,-1^{n+1} \, \cdot \frac{\eta{(n)}}{n} \, = \, log(\frac{\pi}{2})\)
When we do a zeta to eta conversion, we lose log(2), and are adding different logs together to get log (pi/4), which is log (pi) - 2 log(2):
\(
\begin{array}{|c|c|c|c|c|c|c|}
\hline & 0\cdot \frac{ \zeta(1)}{1} & - \frac{1}{2} \cdot \frac{\zeta(2)}{2} & \frac{3}{4} \cdot \frac{\zeta(3)}{3} & -\frac{7}{8} \cdot \frac {\zeta(4)}{4} & \frac{15}{16} \cdot \frac{\zeta(5)}{5} & \dots\\
\hline -log(\frac{9}{8}) & 0\cdot\frac{1}{1 \cdot 1^1} & - \frac{1}{2} \cdot\frac{1}{2 \cdot 1^2} &\frac{3}{4} \cdot \frac{1}{3 \cdot 1^3}& -\frac{7}{8} \cdot \frac{1}{4 \cdot 1^4}&\frac{15}{16} \cdot \frac{1}{5 \cdot 1^5}&\dots\\
\hline -log (\frac{5^2}{24}) & 0\cdot\frac{1}{1 \cdot 2^1} & - \frac{1}{2} \cdot\frac{1}{2 \cdot 2^2} &\frac{3}{4} \cdot \frac{1}{3 \cdot 2^3}&-\frac{7}{8} \cdot \frac{1}{4 \cdot 2^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 2^5}&\dots\\
\hline -log (\frac{7^2}{48}) &0\cdot \frac{1}{1 \cdot 3^1} &- \frac{1}{2} \cdot\frac{1}{2 \cdot 3 ^2} &\frac{3}{4} \cdot \frac{1}{3 \cdot 3^3}& -\frac{7}{8} \cdot \frac{1}{4 \cdot 3^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 3^5} &\dots\\
\hline -log (\frac{9^2}{80}) &0\cdot \frac{1}{1 \cdot 4^1} &- \frac{1}{2} \cdot \frac{1}{2 \cdot 4^2} &\frac{3}{4} \cdot \frac{1}{3 \cdot 4^3}&-\frac{7}{8} \cdot \frac{1}{4 \cdot 4^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 4^5} &\dots\\
\hline- log (\frac{11^2}{120}) &0\cdot \frac{1}{1 \cdot 5^1} & - \frac{1}{2} \cdot\frac{1}{2 \cdot 5^2} &\frac{3}{4} \cdot \frac{1}{3 \cdot 5^3}& -\frac{7}{8} \cdot \frac{1}{4 \cdot 5^4}& \frac{15}{16} \cdot \frac{1}{5 \cdot 5^5} &\dots\\
\hline \dots &\dots & \dots&\dots &\dots&\dots&\dots\\
\hline
\end{array}
\)
Gets us the interesting identities:
\( pi = e^ { 2\, log(2) \, + \sum\limits_{s=2}^\infty \, \frac{\zeta(s)}{s} \cdot (-1)^{s+1} \cdot (1-2^{1-s})\)
And the "log shift function" gets us another interesting identity:
\( \sum\limits_{s=2}^\infty \, \frac{\zeta(s)}{s} \cdot (-1)^{s+1} \cdot (1-2^{1-s}) = \sum\limits_{s=2}^\infty \, (\frac{\zeta(s)-1}{s}) \cdot (1-2^{1-s}) \)
Or....
\( \sum\limits_{s=2}^\infty \, \frac{\zeta(s)}{s} \cdot (-1)^{s+1} = \sum\limits_{s=2}^\infty \, \frac{\zeta(s)-1}{s} \)
Didn't connect it to pi yet.
