# Infinite sum of powerseries likely converges to a powerseries with rational coefficients and has then a simple generating function... proof?

Background: In a discussion in the "tetration-forum" a term $$\log(x-L)/L+\log(x-L^*)/L^*$$ occured, where $$L$$ and $$L^*$$ mean the (complex) primary fixpoint (and its conjugate) of the $$\exp()$$-function. Because the conjugate values occur symmetrically in the sum of the two formal powerseries, the result is a formal powerseries with real coefficients, and we could as well write $$T(x) = 2 \cdot \Re (\log(x - L)/L)$$ where $$\Re()$$ means the real part of its argument.

Problem: Because the exponential function has infinitely many complex fixpoints I thought what would a formal powerseries $$S(x)$$ look like, when it is the sum off all such $$T()$$ - or: when all that fixpoints are taken into account?

Let's write $$L_k$$ for the $$k$$'th fixpoint, with $$L_1 \approx 0.318131505205 + 1.33723570143 \, î$$ , then $$T_k(x)=2 \cdot \Re (\frac {\log(x-L_k)}{L_k}) \tag 1$$ (here, using the convention of W|A, we get the $$k$$'th fixpoint by $$\small {L_k = \exp(-\text{productlog}(-k,-1))}$$).

Now the sum of all that $$T_k(x)$$ (taken as formal power series) gives another formal powerseries $$S_n(x) = \sum_{k=1}^n T_k(x) = s_{n:0} + s_{n:1} x + s_{n:2} x^2 + \cdots \tag 2$$ where -heuristically- it seems, that for $$n \to \infty$$ the values of $$s_{n:0}$$ diverges but all other coefficients $$s_{n:k}$$ converge to a finite value, and moreover, to rational values.

Heuristics: Here are the leading coefficients for some $$S_n(x)$$ up to $$n=2^{16}-1$$

  n    s_{n:0}          s_{n:1}          s_{n:2}          s_{n:3}        s_{n:4}            ....
-------------------------------------------------------------- ----------------------------
1  -2.44695072007  0.945130773416  0.248253690529  -0.111008639310  -0.0937330420633
3  -2.93357528303  0.982250645218  0.249909831374  -0.111105628916  -0.0937498749582
7  -3.29668141804  0.992566300100  0.249991660549  -0.111110673480  -0.0937499978358
15  -3.63438202949  0.996574875354  0.249999033336  -0.111111067412  -0.0937499999458
31  -3.96777440601  0.998353868014  0.249999876694  -0.111111106232  -0.0937499999984
63  -4.30277816467  0.999192862471  0.249999983621  -0.111111110535  -0.0937499999999
127  -4.64069484715  0.999600341105  0.249999997790  -0.111111111041  -0.0937500000000
255  -4.98130134424  0.999801140528  0.249999999701  -0.111111111102  -0.0937500000000
511  -5.32397016248  0.999900812031  0.249999999960  -0.111111111110  -0.0937500000000
1023  -5.66808038151  0.999950466396  0.249999999995  -0.111111111111  -0.0937500000000
2047  -6.01314198265  0.999975248290  0.249999999999  -0.111111111111  -0.0937500000000
4095  -6.35880718549  0.999987627918  0.250000000000  -0.111111111111  -0.0937500000000
8191  -6.70484432699  0.999993814902  0.250000000000  -0.111111111111  -0.0937500000000
16383  -7.05110555754  0.999996907687  0.250000000000  -0.111111111111  -0.0937500000000
32767  -7.39749940344  0.999998453902  0.250000000000  -0.111111111111  -0.0937500000000
65535  -7.74397059311  0.999999226966  0.250000000000  -0.111111111111  -0.0937500000000
....


Conjecture 1: coefficients $$s_{n:k}$$ are rational in the limit for $$n \to \infty$$ (except for $$s_{n:0}$$ which diverges to $$-\infty$$)

If we assume, that indeed that coefficients converge to rational values, we can arrive at a sequence of integer coefficients by a very simple rational scaling:

$$\begin{array} {} \lim_{n \to \infty} S_n(x)= S(x)&= s_0 &+ 1\cdot x + 1\cdot \frac{x^2}{2!}\frac12 − 2\cdot \frac{x^3}{3!}\frac13 − 9\cdot \frac{x^4}{4!}\frac14 \\ && + 6\cdot \frac{x^5}{5!}\frac15 + 155\cdot \frac{x^6}{6!}\frac16 + 232\cdot \frac{x^7}{7!}\frac17 + ... \end{array} \tag 3$$

Coefficients seem to be known:
The miraculous database of integer-sequences, OEIS, knows this coefficients $$[1,1,-2,-9,6,155,232, \ldots]$$ saying they have the exponential generating function (which I modify here slightly for my purposes): $$\begin{array} {} U(x) = \frac{\log(1- x\exp(-x))}x &= -1 &+ 1\cdot x\frac12 + 1\cdot \frac{x^2}{2!}\frac13 − 2\cdot \frac{x^3}{3!}\frac14 − 9\cdot \frac{x^4}{4!}\frac15 \\ &&+ 6\cdot \frac{x^5}{5!}\frac16 + 155\cdot \frac{x^6}{6!}\frac17 + ... \end{array} \tag 4$$

Here the coefficient at the constant term is $$u_0=-1$$ and is likely different to the value of $$s_0$$ which is a result of a likely divergent series.

I've a simple modification of the $$U()$$-function which matches then the conjectured rational coefficients of $$S(x)$$ even by the indexes:

$$U_1(x)={\small{\exp(x)-1 \over \exp(x)-x} }\tag {5.a}$$ $$U_2(x) = \int { \small{\frac{U_1(x)}x}} dx + s_0 \tag {5.b}$$ $$\qquad \qquad$$ $$U_1$$ is a reformulation of the derivative of $$U(x)$$ and $$U_2(x)$$ a termwise integration

Then Pari/GP gives me the following powerseries:

U_2(x)= s0 + x + 1/4*x^2 - 1/9*x^3 - 3/32*x^4 + 1/100*x^5 + 31/864*x^6 + 29/4410*x^7 - 63/5120*x^8 - 2087/326592*x^9 + 39593/12096000*x^10 + 45973/12196800*x^11 - 146387/522547200*x^12 - 10264123/5782233600*x^13 - 2678759/6258954240*x^14 + 833302651/1225944720000*x^15 + 46063312597/111588212736000*x^16 + O(x^17)


Conjecture 2: The coefficients of the limiting formal powerseries of $$S(x)$$ are the same as that of $$U_2(x)$$

This seems to be a nice coincidence - if the assumption (conjecture 1) of convergence of the coefficients in $$S(x)$$ to that rational values holds. I chewed a bit on how to approach a proof, but didn't have a promising idea yet.

Q1: how could this apparent coincidence of the limit of the sum-of-powerseries in $$S(x)$$ with the coefficients in $$U(x)$$ (or better $$U_2(x)$$) be proved?
Q2: can the value of $$s_0$$ be expressed by a regularized summation?

• The directly relevant posting is math.eretrandre.org/tetrationforum/… May 17, 2021 at 15:14
• Thank you for posting such questions. Unfortunately Im not as skilled as the people at the tetration forum or the better ones here. But when I have time I might look into it. Congrats with $10 \pi k$ reputation :)
– mick
May 24, 2021 at 21:37
• Im not sure but I know there exists math for sums of functions taken over the zero's of other functions. See for instance at the second half of this : mathworld.wolfram.com/RiemannZetaFunctionZeros.html
– mick
May 27, 2021 at 0:58
• @mick - thanks for the link, nice compilation there! I've met infinite sums involving roots from time to time, but never delved into it deeper. I'll be happy if this here shall have any significance for J.D.Fox' analysis (in the tetration-forum) of his slog()-powerseries who used only the sum with one single fixpoint: it seems the powerseries $\text{res}(x)=\text{slog}(x)-S(x)$ might be entire... and $\text{slog}(x)=\text{res}(x)+S(x) = \text{res}(x) + U_2(x)$ might then be evaluable to higher precision... May 27, 2021 at 6:23
• @Mick - hi mick, thanks for your engagement. I made an error - usually I check the hypothese of "entireness" on base of increasing ratio of consecutive coefficients of the taylor series. Don't know what I did wrong the day I'd written this: recalculation doesn't support such hypothese: "entireness" no more conjectured! --- The link to the tetration-forum is in the beginning of the post. May 28, 2021 at 6:53