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

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 $\forall s \in \mathbb{C} \setminus \{1\}, Re(s_0 ) > 0$ : $$\zeta(s) = \zeta(s_0)+\frac{1}{s-s_0}\sum_{n=1}^\infty \frac{(s_0-s)^n}{n!}\sum_{m=1}^\infty m^{-s_0} \ln^{n-1} (m) ((s_0-1)\ln (m)-n)$$

  • 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
  • @user243301 Are you sure ? Can you write all the steps for proving you are right ? – reuns Dec 7 '16 at 12:15
  • 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
  • 1
    @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
  • Yes I was thinking in it, few seconds ago, now I understand, sorry me. – user243301 Dec 7 '16 at 12:24

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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