# On the Maclaurin expansion of the Riemann zeta function and a related sequence.

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.

• $\sum_{i=1}^\infty{\ln(i)^n}$ diverges for all $n\ge 0$. So how do you interpret you power series? – gammatester Nov 25 '16 at 9:20
• Referring to the first formula, just let the first sum grow faster than the second sum. – tyobrien Nov 25 '16 at 11:52
• @gammatester You are misreading it. The sum:$$\sum_{n=0}^\infty\left(\frac{(-1)^n}{n!}\sum_{i=1}^m[\ln(i)]^n\right)s^n$$converges for each $m$ (and appropriate $s$), and the limit as $m\to\infty$ exists. For obvious reasons, we cannot take the limit inside the sum. In the same manner,$$\frac12=\lim_{x\to-1^+}\frac1{1-x}=\lim_{x\to-1^+}\sum_{k=0}^\infty x^k\ne\sum_{k=0}^\infty(-1)^k\text{ diverges}$$ – Simply Beautiful Art Jul 23 '17 at 18:38
• @simply-beautiful-art: My comment deals with the original questions from Nov. 2016, where the inner sum was $\sum_{i=1}^\infty$, the upper bound $m$ appears in the edit-version from Feb, 2017. – gammatester Jul 23 '17 at 20:59
• @gammatester Ah, okay – Simply Beautiful Art Jul 23 '17 at 21:08

$\zeta(s)$ 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 $\forall s \in \mathbb{C} \setminus \{1\}, Re(s_0 ) > 0$ : $$\zeta(s) = \zeta(s_0)+\frac{1}{s-s_0}\sum_{n=1}^\infty \frac{(s_0-s)^n}{n!}\sum_{m=1}^\infty m^{-s_0} \ln^{n-1} (m) ((s_0-1)\ln (m)-n)$$
• 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 $(-1)^{n-1}$. – user243301 Dec 7 '16 at 10:38
• 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. – user243301 Dec 7 '16 at 12:16
• @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 $(s_0-1)h_m^{(n)}(s_0) = (s_0-1)(-1)^{n}m^{-s_0}\ln^{n}(m)$ – reuns Dec 7 '16 at 12:22