Finding a solution to $\sum\limits_{n=1}^{n=k} \frac{1}{n^s}=0$

Question

Finding ONE solution to: $$\sum\limits_{n=1}^{n=k} \frac{1}{n^s}=0$$

can apparently be done by iterating the following formula:

$$\Large s(m+1)=\frac{\log \left(-\frac{1}{\sum _{n=1}^{k-1} \frac{1}{n^{s(m)}}}\right)}{\log (k)}+\frac{2 i \pi (k-1)}{\log (k)}$$

where $s(1)=0$.

That can be achieved with the following Mathematica program:

(*Mathematica 8 start*)Clear[nn, cc, s, x, y];
Print["k can be varied to some (not all) integers greater than or equal to 2:"]
k = 5;
cc = 4000;
s = 0;
Do[s = (2 I \[Pi]*(k - 1))/Log[k] + 
   N[Round[Log[1/(-Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^120]/10^120, 
    120], {i, 1, cc}]
Print["x"]
x = s
Print["y"]
y = s = (2 I \[Pi]*(k - 1))/Log[k] + 
   N[Round[Log[1/(-Sum[1/n^s, {n, 1, k - 1}])]/Log[k]*10^120]/10^120, 
    120]
Print["Because the result below is zero, x and y are solutions to \ 
 the equation in the question."]
sumx = Sum[1/n^(x), {n, 1, k}]
sumy = Sum[1/n^(y), {n, 1, k}]
RealDigits[Im[x]][[1]][[1 ;; 80]]
(*end*)

So for example: If we want to find ONE solution to:

$$\sum\limits_{n=1}^{n=5} \frac{1}{n^s}$$

we set $k=5$ in the program. and iterate:

$$\Large s(n+1)=\frac{\log \left(-\frac{1}{\sum _{n=1}^{5-1} \frac{1}{n^{s(n)}}}\right)}{\log (5)}+\frac{2 i \pi (5-1)}{\log (5)}$$

The zero it spits out is:

-1.8514068355116267824936466917494306073203302851205686987125101058514\ 1135175863746506007983472052570016587949681000178352 + 16.838573488618465817880974439653507217425616288174917191378544458658\ 96920254534471600967789655242655156731483566832719867 I

which is a complex number.

Plugging this number into

$$\sum\limits_{n=1}^{n=5} \frac{1}{n^s}$$

we get:

0.*10^-118 + 0.*10^-118 I which is practically zero.

Can you prove this, or did someone already prove it yesterday since I posted a related problem and its conjectured solution earlier this week?

The starting point was this Mathematica command: Reduce[1 + 1/2^a + 1/3^b == 0, {a, b}]

Which gives: $$C[1]\in \text{Integers}\&\&2^a \left(1+2^a\right)\neq 0\&\&b==\frac{2 i \pi C[1]}{\text{Log}[3]}+\frac{\text{Log}\left[\frac{1}{-1-2^{-a}}\right]}{\text{Log}[3]}$$

Update
The formula only works for certain values of k: k=4, k=5, k=7, k=8, k=10, k=11, k=16

but not k=3, not k=6 not k=9, not k=12. I don't know what the pattern is. Sofar multiples of 3 needs to be avoided.

Answer 1

There are MANY solutions to those equations. They are all located in a vertical band, and I'm almost sure there are $\frac {h (\log k - \log 1)}{2\pi} + O(1)$ solutions in a strip of height $h$.

To be able to find a specific solution with your recursion, first you have to pick a branch cut of $\log$ where you can find it (if you pick a cut randomly there is one solution on average, sometimes more, sometimes none at all) ; and then your solution needs to be attractive for the recursion to have a chance to converge to it.

The derivative at a solution $s$ is given by $-(\sum_{n=1}^{k-1} \log n .n^{-s})/\log k. k^{-s}$, and you need this to have modulus less than $1$. Experimentally, when $\Re (s)$ is smaller the less likely this is to happen because the denominator gets too small. It definitely looks like a positive proportion of the solutions are not attractive.

Overall this doesn't strike me as a good way to find the solutions (it misses a large number of them), and having to make a branch cut for $\log$ makes it very clunky and annoying to study.