I attached my best guess. I hope it helps.

Disappointed with Mathematica for this type of problem. I tried all I could've tried to no avail, Mathematica doesn't provide any reasonable solution using NSolve, not even an approximate solution, it's always empty $\rightarrow$ {}. Mathematica let me down.

Here's a link for the article where I wanted to use that: Graph with mysterious roots

you guys are making the same mistake as me when I first started, it would explode out to infinity leaving me puzzled. It's not gonna work the traditional way, you have to use ListPlot.

Here's the code for you to try:

 $MaxExtraPrecision = 1000; m = 9; k = 100*m; Clear[f, g]; 
f[z_] := N[-(2/Cos[\[Pi] z]) \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(Max[16  z, 12]\)]\(
\*SuperscriptBox[\((\(-1\))\), \(i\)] 
\*SuperscriptBox[\((2  \[Pi]\ z)\), \(2  i\)] \(
\*UnderoverscriptBox[\(\[Sum]\), \(j = 1\), \(i\)]
\*FractionBox[\(
\*SuperscriptBox[\((\(-1\))\), \(j\)] 
\*SuperscriptBox[\((2  \[Pi])\), \(\(-2\) j\)] 
\*SuperscriptBox[\(Zeta[2  j]\), \(-1\)]\), \(\((2  i + 1 - 
          2  j)\)!\)]\)\)\), 20]; \[Mu] = 
 Table[{n = m (q - 1)/(k - 1), f[n]}, {q, 1, k}]; g2 = 
 ListPlot[{\[Mu]}, Joined -> True, ImageSize -> Large, 
  PlotStyle -> {Thickness[Tiny]}, 
  GridLines -> {Table[i, {i, 1, m}], Table[f[i], {i, 1, m}]}]

Of course.

As I said before, as a rule of thumb I believe setting $M$ to $16 x_0$ gives a very accurate convergence for all $x$ such that $|x|\le x_0$. So for example, $M=50$ is good for $x$ up to 3, but to save on processing time I set $M=16x$ (Mathematica truncates that to an integer value on a sum). And for $|x|<1$, I make $M=12$.