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

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]][]
(*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]][], {kkk, 0, 100}], kkk]


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

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

• Anyway, for every integer $k$ and for any complex number $x$, you have that $$(x, x+(2k+1)\pi i)$$ is a trivial solution for your equation. – Crostul Jan 13 '16 at 14:46
• ok, I see. I will try that. – Mats Granvik Jan 13 '16 at 14:50
• For arbitrary $1 \leqslant m \leqslant k$, the function $$f_{m,k}(z) = \sum_{n = m}^k \frac{1}{n^z}$$ is an entire transcendental function. By Picard's little theorem, there is at most one complex value that is not attained by $f_{m,k}$. So there are at most countably many $x\in \mathbb{C}$ such that there is no $y\in \mathbb{C}$ with $f_{m,k}(y) = - f_{m,k}(x)$. Hence you can always find pairs $(x,y)$ with $f_{m,k}(x) + f_{m,k}(y) = 0$. – Daniel Fischer Jan 13 '16 at 15:45
• I'm not too sure what to make of this question. Are you asking if there still is an attractive $2$-cycle when you change the bounds of the sums ? – mercio Jan 14 '16 at 1:48
• @Crostul: I don't get you: $n^{x + (2k+1) \pi \rm i} = {\rm e} ^{x + (2k+1) \pi {\rm i} \ln n} \ne - n^x$, so the terms corresponding to the same $n$ from the two sums do not cancel. – Alex M. Jan 14 '16 at 12:33