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I'm studying the Maclaurin series for the Riemann zeta function. I got that for $\Re(s)\ge1$ we have $$\zeta(s)=\lim\limits_{m\to\infty}\sum_{n=0}^\infty\left(\frac{(-1)^n}{n!}\sum_{i=1}^m{\ln(i)^n}\right)s^n$$ if we allow $0^0=1$

By looking at patterns, I formulated a guess for what the continuation would be for $0\lt{\Re(s)}\lt1$. (Again, allow $0^0=1$).

$$\zeta(s)=\lim\limits_{m\to\infty}\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{a(k)}{(n-k)!}\ln(2)^k\sum_{i=1}^m(-1)^{(n-k+i)}\ln(i)^{(n-k)}\right)s^n$$ $a(k)$ is a sequence of rational numbers. Starting at $k=0$ they go $$\{1,2,3,\frac{13}{3},\frac{25}{4},\frac{541}{60},\frac{1561}{120},\frac{47293}{2520},\dots\}$$

I noticed that $\frac{a(k-1)}{a(k)}$ gets closer and closer to $\ln2$ for larger and larger $k$. I took a guess for the sake of computation that $$a(k)=\prod_{i=0}^k\frac{\left[\frac{i!}{\ln(2)^{(i+1)}}\right]}{\left[\frac{i!}{\ln(2)^i}\right]}$$ where the brackets notate the nearest integer function. Please check me on my formulas and on this sequence and help me find a better formula to define the sequence.

Edit (1/30/2021)

I came back to this just to mess with it and have some new formulae that I would like to add to the body of this post just for records.

I believe that the function is actually the Polylogarithm, $a(x)=L_{-x}\left(\frac12\right)$ This allows us to derive $$\zeta(s)=\sum_{n=1}^{\infty}\left(-1\right)^{n}\sum_{k=0}^{\infty}\frac{s^{k}}{k!}\sum_{i=1}^{\infty}\frac{\left(i\ln2-\ln n\right)^{k}}{2^{i}}$$ for $0\le\mathbb{R}(s)<1$.

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  • $\begingroup$ Referring to the first formula, just let the first sum grow faster than the second sum. $\endgroup$
    – tyobrien
    Commented Nov 25, 2016 at 11:52

1 Answer 1

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You need to read a complex analysis course

$\zeta(s)$ is meromorphic with a pole at $s=1$, so in every case its Taylor series diverges at $s=1$.

Now $F(s) = (s-1)\zeta(s)$ is entire, so that $$\forall s,s_0, \qquad \qquad F(s) = \sum_{n=0}^\infty \frac{F^{(n)}(s_0)}{n!} (s-s_0)^n$$ And for $Re(s_0) > 1$, with $G(s) = s-1$ : $$F^{(n)}(s_0)=\sum_{k=0}^n {n \choose k} \zeta^{(n-k)}(s_0)G^{(k)}(s_0)=(s_0-1)\zeta^{(n)}(s_0)+n\zeta^{(n-1)}(s_0)$$ $$ = (-1)^{n-1}\sum_{m=1}^\infty m^{-s_0} \ln^{n-1} (m) (n-(s_0-1)\ln (m))$$ Whence $\forall s \in \mathbb{C} \setminus \{1\}, Re(s_0 ) > 0$ : $$\zeta(s) = \zeta(s_0)+\frac{1}{s-s_0}\sum_{n=1}^\infty \frac{(s_0-s)^n}{n!}\sum_{m=1}^\infty m^{-s_0} \ln^{n-1} (m) ((s_0-1)\ln (m)-n)$$

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  • $\begingroup$ I belive that there are minor typos, since your second identity should be $$F^{(n)}(s_0)= (-1)^{n-1}\sum_{m=1}^\infty m^{-s_0} \ln^{n-1} (m) (n+(s_0-1)\ln (m)),$$ and in your final result I believe that you forgot the sign $(-1)^{n-1}$. $\endgroup$
    – user243301
    Commented Dec 7, 2016 at 10:38
  • $\begingroup$ @user243301 Are you sure ? Can you write all the steps for proving you are right ? $\endgroup$
    – reuns
    Commented Dec 7, 2016 at 12:15
  • $\begingroup$ THe first claim yes, because you change the sign in the factor $(n+(s_0-1)\ln (m))$. Which I am saying when I say: I believe that there are typos, is that I am almost sure that there are typos. $\endgroup$
    – user243301
    Commented Dec 7, 2016 at 12:16
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    $\begingroup$ @user243301 With $h_m(s) = m^{-s}$ then $(s_0-1)\zeta^{(n)}(s_0)+n\zeta^{(n-1)}(s_0) = \sum_{m=1}^\infty (s_0-1)h_m^{(n)}(s_0)+nh_m^{(n-1)}(s_0)$. Now $h_m^{(n-1)}(s_0) = (-1)^{n-1}m^{-s_0}\ln^{n-1}(m)$ and $(s_0-1)h_m^{(n)}(s_0) = (s_0-1)(-1)^{n}m^{-s_0}\ln^{n}(m)$ $\endgroup$
    – reuns
    Commented Dec 7, 2016 at 12:22
  • $\begingroup$ Yes I was thinking in it, few seconds ago, now I understand, sorry me. $\endgroup$
    – user243301
    Commented Dec 7, 2016 at 12:24

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