Let $\log(1),\log(2),\log(3),\log(4)...\log(n)$ be approximated by fractions generated by the truncated sums:

$k=0$

$c=1$

$$\text{log1}=\sum_{n=0}^k \frac{0}{(1 n+1)^c}=0$$
$$\text{log2}=\sum _{n=0}^k \left(\frac{1}{(2 n+1)^c}-\frac{1}{(2 n+2)^c}\right)=\frac{1}{2}$$
$$\text{log3}=\sum _{n=0}^k \left(\frac{1}{(3 n+1)^c}+\frac{1}{(3 n+2)^c}-\frac{2}{(3 n+3)^c}\right)=\frac{5}{6}$$
$$\text{log4}=\sum _{n=0}^k \left(\frac{1}{(4 n+1)^c}+\frac{1}{(4 n+2)^c}+\frac{1}{(4 n+3)^c}-\frac{3}{(4 n+4)^c}\right)=\frac{13}{12}$$

Letting $k \rightarrow \infty$ makes the sums above converge to the precise values of $\log(n)$ by their Dirichlet generating function:

$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$

The alternating series which is valid in the critical strip can be written as, and is set to zero:

$$\frac{1}{\left(e^{\text{log1}}\right)^s}-\frac{1}{\left(e^{\text{log2}}\right)^s}+\frac{1}{\left(e^{\text{log3}}\right)^s}-\frac{1}{\left(e^{\text{log4}}\right)^s}+...(-1)^{(n+1)}\frac{1}{\left(e^{\log(n)}\right)^s}=0$$

The truncated alternating series:

$$\sum_{n=1}^{n=4}(-1)^{(n+1)}\frac{1}{\left(e^{\log(n)}\right)^s}=0$$ is solved for $s$ in Mathematica 8.0.1 with the program:

```
(*start*)
Clear[log1, log2, log3, log4, n, k, s];
c = 1;
k = 0;
log1 = Sum[0/(1*n + 1)^c, {n, 0, k}];
log2 = Sum[1/(2*n + 1)^c - 1/(2*n + 2)^c, {n, 0, k}];
log3 = Sum[1/(3*n + 1)^c + 1/(3*n + 2)^c - 2/(3*n + 3)^c, {n, 0, k}];
log4 = Sum[
1/(4*n + 1)^c + 1/(4*n + 2)^c + 1/(4*n + 3)^c - 3/(4*n + 4)^c, {n,
0, k}];
$MaxRootDegree = 1000;
Last[Solve[
1/(E^(log1))^s - 1/(E^(log2))^s + 1/(E^(log3))^s -
1/(E^(log4))^s == 0, s]];
FullSimplify[%]
(*end*)
```

which gives the output:

$$\left\{s\to -4 i \pi +\log \left(\text{Root}\left[\text{$\#$1}^{10}-3 \text{$\#$1}^9+3 \text{$\#$1}^8+23 \text{$\#$1}^7+40 \text{$\#$1}^6 \\-2 \text{$\#$1}^5+42 \text{$\#$1}^4+12 \text{$\#$1}^2+1\&,10\right]\right)\right\} \label{1}\tag{$*$}$$

which has the form:

$$\left\{s\to -\text{integer } i \pi +\log \left(\text{polynomial root}\right)\right\}$$

Now the following number also has a similar form:

$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = \\ 14.1347251415...$$
$$14.1347251417...$$
with the value of the actual first Riemann zeta zero `Im[ZetaZero[1]]`

right below it for comparison.
This can't be all coincidence.

**Question:**

How does Mathematica arrive at the output in \eqref{1}$?

In the edit at 19.7.2021 in https://mathematica.stackexchange.com/q/63541/328 I made a guess at the solution for a simpler case which is not the alternating series, but one that also has $1/(e^{\text{fraction}})^s$ as its terms.

`x = N[Root[1 + 12 #1^2 + 42 #1^4 - 2 #1^5 + 40 #1^6 + 23 #1^7 + 3 #1^8 - 3 #1^9 + #1^10 &, 10], 100]; 1 + 12 x^2 + 42 x^4 - 2 x^5 + 40 x^6 + 23 x^7 + 3 x^8 - 3 x^9 + x^10`

which outputs 0 to 93 decimal places:`0.*10^-93 + 0.*10^-93 I`

The number $10$ at the end means the 10-th root of that polynomial. And yes there are many degrees of freedom. $\endgroup$4more comments