The following series seems convergent for all $s\in \mathbb{C}$:

$$\displaystyle f(s):=\sum_{n=1}^\infty \frac{(-1)^n}{(n+s)^{n+s}}$$

The function itself does not appear to have any real or complex zeros, however numerical evidence suggests that all zeros of:

$$f(s) \pm f(1-s)$$

reside on the line $\Re(s)=\frac12$. Their density apparently increases in a quite regular manner.

enter image description here

1) Could this be proven? When using the finite series $\sum_{n=1}^N$, the claim seems to hold for each $N$.

2) Could there exist a functional equation between $f(s)$ and $f(1-s)$? The only relation I found so far is that $f(-1)=f(1)$.

EDIT: A partial result on the second question. Based on Joro's findings below and realising that for integers and half-integers an infinite number of terms could be cancelled out when adding or substracting $f(s)$ and $f(1-s)$, the following reflective formulae hold:

For all integers $s \in \mathbb{Z}$:

$$f(s)+f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$

For all half-integers $s+\dfrac{1}{2}$ with $s \in \mathbb{Z}$:

$$f(s)-f(1-s) +\sum_{n=0}^{2(s-1)} \frac{(-1)^{n}}{(s-n)^{s-n}} =0$$

The non-alternating series $\displaystyle g(s):=\sum_{n=1}^\infty \frac{1}{(n+s)^{n+s}}$ has one formula for integers and half-integers:

$$g(s)-g(1-s) +\sum_{n=0}^{2(s-1)} \frac{1}{(s-n)^{s-n}} =0$$

  • 2
    $\begingroup$ If you want to define this for all $s\in\mathbb C$ directly by the series, you have to be more precise about the powers (a choice of logarithm is involved). $\endgroup$ Oct 4, 2015 at 15:42
  • $\begingroup$ Christian, you are referring to the branch cut of the natural log, right? For the Maple package I used to check convergence for complex values, this is the principle branch between $\pi$ and $-\pi$. $\endgroup$
    – Agno
    Oct 4, 2015 at 16:14
  • $\begingroup$ Yes, $(n+s)^{n+s}=\exp ((n+s)\log (n+s))$, and you have to say which value of $\log$. $\endgroup$ Oct 4, 2015 at 16:44
  • $\begingroup$ I doubt my computation is correct, but one should expect some relation like $f(1-s)=-f(-s)+\dfrac{1}{(1+s)^{(1+s)}}$ to hold. $\endgroup$ Oct 4, 2015 at 16:46
  • $\begingroup$ If you always take values from one fixed interval of length $2\pi$ for the imaginary part of the $\log$, then you have no chance of producing a holomorphic $f$. $\endgroup$ Oct 4, 2015 at 16:48

1 Answer 1


EDIT Due to confusing XRay and programming mistake, I erroneously claimed real zeros. Root finding didn't found any zeros off the line.

Very partial answer for functional equation.

For integer $s$, $\{1,f(s),f(1-s)\}$ appear linearly dependent over the rationals with high precision.

Let $K=f(s),K_2=f(1-s),K_3=1/s^s,K_4=(K-K_2+K_3)^2$.

$$s=2, 1+4K+4K_2=0$$ $$s=3, 131-108K-108K_2=0$$ $$s=4, 36059+ 6912K+ 6912K_2=0$$ $$s=5, -695877463+ 21600000K+ 21600000K_2=0$$ $$s=6, 168087904001+ 583200000K+ 583200000K_2=0$$ $$s=7, -1639334733641495543+ 480290277600000K+ 480290277600000K_2=0$$

And in addition for $s=3/2, K^{2} - 2 \, K K_{2} + K_{2}^{2} + 2 \, K K_{3} - 2 \, K_{2} K_{3} + K_{3}^{2} - 2=0$.

...And with the help of Agno's observation, $s=5/2,8503056K_4^4 - 61096032K_4^3 + 202411224K_4^2 - 171663192K_4 + 252460321$

I suspect for $s$ half integer there is non-trivial algebraic dependency.

  • $\begingroup$ Really nice, Joro! The coefficients 4, 108, 6912,... seem to follow oeis.org/A107048 , so the next one for $s=6$ would be $\pm$2332800000. The constants at the beginning don't seem to appear in a Sloane integer series. $\endgroup$
    – Agno
    Oct 5, 2015 at 7:22
  • $\begingroup$ @Agno for $s=6$ I get: $$168087904001+ 583200000*f(6)+ 583200000*f(1-6)=0$$ which is 4 times of your conjecture. $\endgroup$
    – joro
    Oct 5, 2015 at 7:43
  • $\begingroup$ Thanks Joro. Note that I started experimenting with the function $g(s):=\sum_{n=1}^\infty \frac{1}{(n+s)^{n+s}}$, but that also induces a single complex pair of zeros off the critical line. The alternating version $f(s)$ doesn't seem to have any complex zeros off the line and i.m.o. had therefore more 'beauty'. Also tried to find a pattern linking integer values $g(n)$ and $g(1-n)$ and I believe it has exactly the same 'weights' ($b$) as you found for $f(s)$ however the initial offset values ($a$) are different: $1,9,139,36571,-468908713,136638677249$ for $n=1..6$ and $a+b*g(n)-b*g(1-n)$. $\endgroup$
    – Agno
    Oct 5, 2015 at 13:39
  • $\begingroup$ @Agno I consider significantly changing the question after a counterexample bad practice IMHO. $\endgroup$
    – joro
    Oct 5, 2015 at 14:50
  • $\begingroup$ Joro, I am sorry when I have upset you by making this change. No bad intentions here and I made explicit comments about the change in the edit reason as well as in the OP text. Having said that: are you absolutely sure the graph with the real zeros is correct? I have tried to replicate it and it keeps looking differently with no zeros. For the function $f(s)-f(1-s)$ you would expect zeros on the positive real axis if they show up on the negative axis as well (as in your graph). I get imaginary values (I guess Maple makes a certain branch cut here) for values of $s<-1$. $\endgroup$
    – Agno
    Oct 5, 2015 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.