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.



Are the Riemann zeta zeros of the form $-\text{integer } i \pi +\log \left(\text{polynomial root}\right)$?
The actual question asked here is a *Mathematica* question, so this is not really the right site for it, but here goes. Suppressing the extraneous notation in $c$ and $k$, the expression in $s$ is
$$
-e^{-13 s/12}+e^{-5 s/6}-e^{-s/2}+1
$$
which is a polynomial of degree 13 in $x=e^{-s/12}$:
$$
- x^{13} + x^{10} - x^6+1=-(x-1)(x^2+x+1)(x^{10}+x^3+1)
$$
The Mathematica code 'Solve' tries to find roots of this polynomial in $x$, and the $\log$ expressions will then solve for $s$. (I suspect the OP is showing output from a different choice of $c$ and/or $k$, because already I'm seeing something different.). The first two factors are solvable by radicals of course. The OP has asked for the 'last' solution. To avoid introducing new variables which may conflict with user defined variables, it is merely expressed as Root[1#^10+1#^3+1&,10]. Here 1# is just the variable name (Mathematica can do this in more than one variable - the others would be 2#, 3#, etc.). The & at the end is just a way of saying'Think of this as a function of 1#' - a 'pure function' in the language of the *Mathematica* documentation . The 10 indicated the last root.




---


Addendum: the -integer $\pi i$ in the title is misleading of course. the questions is not about the zeros of $\zeta(s)$, but the imaginary parts of the zeros, which are real numbers. The $\exp(\text{integer}\pi i)$ is just $\pm 1$, and Mathematica is just trying to give the most general form of the solution when one uses Log to solve for $s$.


