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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...

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  • 2
    $\begingroup$ Your first line doesn't make sense. Did you mean finding the zeros of $F(s)=\sum_{n=1}^3 n^{-s}+ 3^{1-s}/(s-1)$ ? If so the first step is to find $a,b$ such that for $\Re(s) \not \in [-a,b]$ then $F(s) \ne 0$ $\endgroup$ – reuns Mar 24 at 13:30
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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.


Contour Plots of Re(F(s))=0 and Im(F(s))=0

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).


Contour Plots of Re(G(s))=0 and Im(G(s))=0 with k=3

Figure (2): Contour Plots of $\Re(G(s))=0$ and $\Im(G(s))=0$ in blue and orange respectively with $k=3$


Contour Plots of Re(G(s))=0 and Im(G(s))=0 with k=100

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 $\Delta_k=1$.


Evaluation of G(s) zero associated with rho_4 from k=3 to k=100

Figure (4): Evaluation of $G(s)$ zero associated with $\rho_4$ from $k=3$ to $k=100$


Evaluation of G(s) zero associated with rho_3 from k=3 to k=100

Figure (5): Evaluation of $G(s)$ zero associated with $\rho_3$ from $k=3$ to $k=100$


Evaluation of G(s) zero associated with rho_2 from k=3 to k=100

Figure (6): Evaluation of $G(s)$ zero associated with $\rho_2$ from $k=3$ to $k=100$


Evaluation of G(s) zero associated with rho_1 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):

$$\zeta(s)=\frac{1}{s-1}\sum_{k=1}^\infty\left(\frac{k}{(k+1)^s}-\frac{k-s}{k^s}\right),\quad \Re(s)>0$$

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  • 1
    $\begingroup$ The proof of $\zeta(s)=\lim\limits_{k \rightarrow \infty} \left( \sum\limits_{n=1}^k \frac{1}{n^s}+ \frac{1}{k^{s - 1} \cdot (s - 1)}\right)$ takes 1 line, the proof of your's takes 4 $\endgroup$ – reuns Jul 2 at 21:28

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