The equation I am trying to solve is:
$$\lim\limits_{k \rightarrow 3} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right)=0 \tag{1}$$
The simplest possible analytic continuation of the Riemann zeta function is: $$\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right) \tag{2}$$ $$\mbox{ which appears to be true for }\Re(s)>0$$
So therefore I substituted all $s$ with $x$ except one of them like this: $$\lim\limits_{k \rightarrow 3} \left( \sum\limits_{n=1}^{n=k} \frac{1}{n^x}+ \frac{1}{k^{x - 1} \cdot (s - 1)}\right)=0 \tag{3}$$
Very crude rational approximations of logarithms are:
$\log(1) = 0$
$\log(2) \approx 7/10$
$\log(3) \approx 11/10$
(That is the level of precision I could afford computationally in this case.)
Notice that:
$$\frac{1}{n^x}=\frac{1}{e^{x\log(n)}} \tag{4}$$
and substitute $\log(n)$ with the rational approximations for logarithms above.
Solving $(3)$ in Mathematica we can then write:
Clear[x, s];
Reduce[1/(E^Round[N[Log[1]], 10^-1])^x +
1/(E^Round[N[Log[2]], 10^-1])^x +
1/(E^Round[N[Log[3]], 10^-1])^x +
1/(E^Round[N[Log[3]], 10^-1])^(x - 1)/(s - 1) == 0, x]
This gives 11 Root objects subject to conditions. Picking the first Root object that Mathematica gives, we have:
x == 10 (2 I \[Pi] C[1] +
Log[Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &,
1]])
Latexifying it does not help much, but the changes I would do are to replace $x$ with $s$ and skip the term $2 i \pi c_1$ since I have understood that $c_1$ is an integer that can be zero.
So the equation that needs to be solved is:
s == 10 (Log[
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]])
Dividing by 10:
s/10 == Log[
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]]
Applying the exponential function we would have:
Exp[s/10] ==
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1]
Now this is not solvable in Mathematica so we need a truncated series expansion of $\exp(s)$
$$\exp(s/10) \approx 1+s/10+\frac{(s/10)^2}{2}+\frac{(s/10)^3}{6}$$
So instead:
$$1+\frac{s}{10}+\frac{1}{2} \left(\frac{s}{10}\right)^2+\frac{1}{6} \left(\frac{s}{10}\right)^3=\text{Root}\left[\text{$\#$1}^{11} (s-1)+\text{$\#$1}^4 (s-1)+s+e^{11/10}-1\&,1\right] \tag{5}$$
and this Mathematica can solve:
Reduce[(1 + s/10 + (s/10)^2/2 + (s/10)^3/6) ==
Root[-1 + E^(11/10) + s + (-1 + s) #1^4 + (-1 + s) #1^11 &, 1], s]
giving the first Root object starting as:
(-1 + Root[-1088391168000000000000000000000000000000000 +
362797056000000000000000000000000000000000 E^(11/10) +
544195584000000000000000000000000000000000 #1 +
295679600640000000000000000000000000000000 #1^2 + ...
Is this at all true or is it just overly complicated?
I mean we could use the same minimal analytic continuation of the zeta function and the crude rational approximations of the logarithms and only do the series expansion:
Clear[x, s];
Series[1/(E^Round[N[Log[1]], 10^-1])^s +
1/(E^Round[N[Log[2]], 10^-1])^s + 1/(E^Round[N[Log[3]], 10^-1])^s +
1/(E^Round[N[Log[3]], 10^-1])^(s - 1)/(s - 1), {s, 0, 5}]
This then would give:
$$\left(3-e^{11/10}\right)+\frac{1}{10} \left(-18+e^{11/10}\right) s+\left(\frac{17}{20}-\frac{101 e^{11/10}}{200}\right) s^2+\left(-\frac{279}{1000}-\frac{1699 e^{11/10}}{6000}\right) s^3+\left(\frac{8521}{120000}-\frac{82601 e^{11/10}}{240000}\right) s^4+\left(-\frac{29643}{2000000}-\frac{3968999 e^{11/10}}{12000000}\right) s^5+O[s]^6$$ and so on...
For $k=3$:
$$\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$$
For $s=0$:
$$1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}=0$$
Note for $s=0$, $\sum\limits_{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}=0$ for all values of $k$.
For $F(s)=1+2^{-s}+3^{-s}+\frac{1}{3^{s-1} (s-1)}$, the contour plots of $\Re(F(s))=0$ (blue curve) and $\Im(F(s))=0$ (orange curve) in Figure (1) below suggest the location of the complex roots of $F(s)$ are in a vertical strip centered near the origin. In Figure (1) below, the horizontal axis is the real axis and the vertical axis is the imaginary axis.
Figure (1): Contour Plots of $\Re(F(s))=0$ and $\Im(F(s))=0$ in blue and orange respectively
For $G(s)=\sum _{n=1}^k \frac{1}{n^s}+\frac{1}{k^{s-1} (s-1)}$, the contour plots of $\Re(G(s))=0$ (blue curve) and $\Im(G(s))=0$ (orange curve) in Figures (2) and (3) below illustrate how four of the zeros of $G(s)$ move closer to the first four non-trivial zeta zeros in the upper-half plane (red discrete portion of the plots) as the evaluation limit is increased from $k=3$ in Figure (2) to $k=100$ in Figure (3).
Figure (2): Contour Plots of $\Re(G(s))=0$ and $\Im(G(s))=0$ in blue and orange respectively with $k=3$
Figure (3): Contour Plots of $\Re(G(s))=0$ and $\Im(G(s))=0$ in blue and orange respectively with $k=100$
The following four figures illustrate how the four zeros of $G(s)$ associated with the first four non-trivial zeta zeros in the upper-half plane ($\rho_1$ to $\rho_4$) migrate from their initial values at $k=3$ to their values at $k=100$ in steps of $\Delta_k=1$.
Figure (4): Evaluation of $G(s)$ zero associated with $\rho_4$ from $k=3$ to $k=100$
Figure (5): Evaluation of $G(s)$ zero associated with $\rho_3$ from $k=3$ to $k=100$
Figure (6): Evaluation of $G(s)$ zero associated with $\rho_2$ from $k=3$ to $k=100$
Figure (7): Evaluation of $G(s)$ zero associated with $\rho_1$ from $k=3$ to $k=100$
Another fairly simple formula I've found useful is the following (see Riemann zeta function - Dirichlet series and Chapter 2 of Part II of "Theory of Functions" by Konrad Knopp):
$$\zeta(s)=\frac{1}{s-1}\sum_{k=1}^\infty\left(\frac{k}{(k+1)^s}-\frac{k-s}{k^s}\right),\quad \Re(s)>0$$