Finding a solution to $\sum _{n=1}^{n=k} \frac{1}{n^x}+\sum _{n=1}^{n=k} \frac{1}{n^y}=0$

Question

Scroll down to the update to see what I am meaning. The Mathematica program below finds a solution to the equation: $$\sum _{n=1}^5 \frac{1}{n^x}+\sum _{n=1}^5 \frac{1}{n^y}=0$$ My question is if you can generalize this algorithm or iterated formula to partial sums of any length, not just $n=5$?

The solution I have found is:
$x=-2.30158037691463871425027298453 + 1.11833640189403133864138145682 I$ $y=-2.30158037691463871425027298453 - 1.11833640189403133864138145682 I$

The program below applies/iterates this equation 300 times:

$$s=\frac{\log \left(-\frac{1}{\sum _{n=1}^{\text{N}-1} \frac{1}{n^s}}\right)}{\log (\text{N})}$$

for $N=5$.

In the program the answer has converged when $s$ alternates between the values for $x$ and $y$ above.

Program:

(*Mathematica 8*)
Clear[c, c1, c2, nn]
nn = 5;
cc = 300;
s = -10000 + 10000*I
Do[
 s = N[Round[Log[1/-Sum[1/n^s, {n, 1, nn - 1}]]/Log[nn]*10^30]/10^30, 
   30], {i, 1, cc}]
Print["x"]
x = s
Print["y"]
y = s = N[
   Round[Log[1/-Sum[1/n^s, {n, 1, nn - 1}]]/Log[nn]*10^30]/10^30, 30]
sumx = Sum[1/n^x, {n, 1, 5}]
sumy = Sum[1/n^y, {n, 1, 5}]
Print["Because the result below is zero, x and y are solutions to the \
equation in the question."]
sumx + sumy
(*end*)

---

Update:

The following equation seems always solvable:

$$\sum _{n=2}^{n=k} \frac{1}{n^x}+\sum _{n=2}^{n=k} \frac{1}{n^y}=0$$

by applying/iterating:

$$s=\frac{\log \left(-\frac{1}{\sum _{n=2}^{\text{k}-1} \frac{1}{n^s}}\right)}{\log (\text{k})}$$

for any integer $k \geq 3$

(*Mathematica 8*)
Clear[c, c1, c2, nn];
Print["nn can be varied to any integer greater than or equal to 3:"]
nn = 11
cc = 2000;
s = -4 - I
Do[s = N[Round[Log[1/-Sum[1/n^s, {n, 2, nn - 1}]]/Log[nn]*10^30]/
    10^30, 30], {i, 1, cc}]
Print["x"]
x = s
Print["y"]
y = s = N[
   Round[Log[1/-Sum[1/n^s, {n, 2, nn - 1}]]/Log[nn]*10^30]/10^30, 30]
sumx = Sum[1/n^x, {n, 2, nn}]
sumy = Sum[1/n^y, {n, 2, nn}]
Print["Because the result below is zero, x and y are solutions to the \
equation in the question."]
sumx + sumy
(*end*)

---

Update 2: As pointed out by Daniel Fischer the following is also solvable:

$$\sum _{n=1}^{n=k} \frac{1}{n^x}+\sum _{n=1}^{n=k} \frac{1}{n^y}=0$$

by applying/iterating:

$$s=\frac{\log \left(-\frac{1}{\sum _{n=1}^{\text{k}-1} \frac{1}{n^s}}\right)}{\log (\text{k})}$$

for any integer $k \geq 2$

(*Mathematica 8*)
Clear[nn, cc, s, x, y];
Print["nn can be varied to any integer greater than or equal to 3:"]
nn = 7;
cc = 2000;
s = -4 - I;
Do[s = N[Round[Log[1/(-Sum[1/n^s, {n, 1, nn - 1}])]/Log[nn]*10^80]/
    10^80, 80], {i, 1, cc}]
Print["x"]
x = s
Print["y"]
y = s = N[
   Round[Log[1/(-Sum[1/n^s, {n, 1, nn - 1}])]/Log[nn]*10^80]/10^80, 80]
sumx = Sum[1/n^(x), {n, 1, nn}]
sumy = Sum[1/n^(y), {n, 1, nn}]
Print["Because the result below is zero, x and y are solutions to the \
equation in the question."]
sumx + sumy
RealDigits[Im[x]][[1]]
(*end*)

Edit 13.2.2016: The program below:

(*Mathematica 8*)
Clear[k, m, s, kk, n, aa, b1, t, aaa, kkk, aaa]
Print["k can be varied to any integer greater than or equal to 2:"]
k = 3;
m = 1000;
s = 0;
Monitor[aaa = 
  Table[Table[
    Do[s = (2 I \[Pi]*(kk))/Log[k]*If[Mod[i, 2] == 0, -1, 1] + 
       N[Round[Log[-1/(Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^20]/
         10^20, 20], {i, 1, m}];
    aa = Table[
      s = (2 I \[Pi]*(kk))/Log[k]*If[Mod[i, 2] == 0, -1, 1] + 
        N[Round[Log[-1/(Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^20]/
          10^20, 20], {i, 1, 100}];, {kk, kkk, kkk}];
   (*end*)
   Flatten[
     Position[Chop[Accumulate[Sum[1/n^(aa), {n, 1, k}]]], 
      0]][[1]], {kkk, 0, 100}], kkk]

appears to always find one of the solution pairs $x,y$ to:

$$\sum _{n=1}^3 \frac{1}{n^x}+\sum _{n=1}^3 \frac{1}{n^y}=0$$