Maclaurin Series of the Riemann Zeta

Question

I am high school doing a maths essay on the Maclaurin Series of the Zeta function, but I can't find much. I just wanted to ask how close is my series to the correct Maclaurin function?

$\sum_{n=1}^{\infty} \frac{1}{n^0}$ - $\sum_{n=2}^{\infty} \frac{\ln(n) }{n^0 \cdot 1!}s$ + $\sum_{n=2}^{\infty} \frac{\ln(n)^2 }{n^0 \cdot 2!}s^2$ - $\sum_{n=2}^{\infty} \frac{\ln(n)^3 }{n^0 \cdot 3!}s^3$ $+ ... +$ $(-1)^x\sum_{n=2}^{\infty} \frac{\ln(n)^x }{n^0 \cdot x!}s^x$

Possible duplicate of On the Maclaurin expansion of the Riemann zeta function and a related sequence. — Wraith1995, Jul 23 '17 at 17:23
$\sum_{n=1}^{\infty} \frac{1}{n^0}$ does not converge and your other summations are even more not convergent. — Somos, Jul 23 '17 at 17:47
@Ian_16 Somos means exactly what he says, $\sum_{n=1}^\infty1$ is a divergent sum. — Simply Beautiful Art, Jul 23 '17 at 18:20
@Ian_16 One of the fundamental tools of Calculus is summation of infinite series and it is very important to know if they converge to a definite value or else they don't. What real number do you think your first summation converges to? — Somos, Jul 23 '17 at 18:27
@Somos I just caught on to what you were saying. I was hoping that it would converge to 0.5. — Ian_16, Jul 23 '17 at 18:54
@Ian_16 0.5!? Surely you at least meant $\zeta(0)=-1/2$? — Simply Beautiful Art, Jul 23 '17 at 19:03
If you are in high school and if your class knows complex numbers, then you should explain first the analytic continuation of the geometric series : $\sum_{n=0}^\infty z^n$ converges for $|z| < 1$ to $\frac{1}{z-1}$ which is well-defined for $z \in \mathbb{C} \setminus {0}$. — reuns, Jul 23 '17 at 19:09
@reuns no we haven't but if I have to know these things, I'll gladly do it. Anything else? — Ian_16, Jul 23 '17 at 19:15
@Ian_16 His point is that in order to properly understand the Riemann zeta function, you need to understand analytic continuation. For example, we do not calculate $\zeta(-1)$ by adding up all the natural numbers... do you know how $\zeta(-1)$ is properly defined? — Simply Beautiful Art, Jul 23 '17 at 19:18
I meant $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ for $z \in \mathbb{C},|z|< 1$. Do you know complex numbers, for example $e^{i \pi} = -1, e^{i \pi/2} = \sqrt{-1}$ ? @Ian_16 — reuns, Jul 23 '17 at 19:22
@Ian_16 : Then $\frac{1}{1-z}, z \in \mathbb{C} \setminus {1}$ is the analytic continuation of $\sum_{n=0}^\infty z^n$. For $\zeta(z)$ it works the same way : $\zeta(z) = \sum_{n=1}^\infty n^{-z}$ is well-defined only for $\Re(z) > 1$, but it has an analytic continuation to $z \in \mathbb{C} \setminus {1}$. This is the most important point with the Riemann zeta function. You should probably play with Mathematica to understand this concretely. — reuns, Jul 23 '17 at 19:27

score 2 · Answer 1 · answered Jul 23 '17 at 19:02

You probably meant to say

$$\zeta(s)=\sum_{n=0}^\infty\frac{\zeta^{(n)}(0)}{n!}s^n$$

But note that:

$$\zeta(0)=-\frac12\ne\sum_{n=1}^\infty1=\infty\\\zeta'(0)=-\frac12\ln(2\pi)\ne\sum_{n=1}^\infty\ln(n)=\infty\\\zeta''(0)=\dots\ne\sum_{n=1}^\infty\ln^2(n)=\infty\\\vdots$$

If one so wishes for such a series expansion, it is possible to use the relationship to the Dirichlet eta function:

$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)$$

where we have

$$\eta(s)=\lim_{x\to-1^+}\sum_{k=1}^\infty\frac{x^{k+1}}{k^s}$$

and more generally,

$$\eta^{(n)}(s)=\lim_{x\to-1^+}\sum_{k=1}^\infty\frac{x^{k+1}\ln^n(k)}{k^s}$$

$$\eta(s)=\lim_{x\to-1^+}\sum_{n=0}^\infty\frac{s^n}{n!}\sum_{k=1}^\infty x^{k+1}\ln^n(k)$$

Likewise, it is easy to see that

$$\frac1{1-2^{1-s}}=\sum_{k=0}^\infty2^{k(1-s)}=\sum_{n=0}^\infty\frac{s^n}{n!}\sum_{k=0}^\infty(-k\ln(2))^n2^k$$

Take the Cauchy product and you get

$$\zeta(s)=\sum_{n=0}^\infty a_ns^n$$

where,

$$a_n=\lim_{x\to-1^+}\sum_{j=0}^n\sum_{k=1}^\infty\frac{x^{k+1}\ln^j(k)}{j!}\sum_{l=0}^\infty\frac{(-l\ln(2))^{n-j}2^l}{(n-j)!}$$

I would mention $(s-1) \zeta(s)$ is entire, and write its Taylor series at say $s=2$ — reuns, Jul 23 '17 at 19:08
@reuns Tbh, I'd personally prefer to take the Laurent expansion about $s=1$. :| Would save me the trouble of all these messy expressions. — Simply Beautiful Art, Jul 23 '17 at 19:10
The expansion at $s=1$ is more interesting but needs another expression for $\zeta$ — reuns, Jul 23 '17 at 19:11

score 0 · Answer 2 · edited Jun 12 '20 at 10:38

If you understood what are complex numbers and series,

$\frac{1}{1-z}, z \in \mathbb{C} \setminus \{1\}$ is the analytic continuation of $\sum_{n=0}^\infty z^n$.

For $\zeta(z)$ it works the same way : $\zeta(z) = \sum_{n=1}^\infty n^{-z}$ is well-defined only for $\Re(z) > 1$, but it has an analytic continuation to $z \in \mathbb{C} \setminus \{1\}$.

This is the most important point with the Riemann zeta function. You should probably play with Mathematica or WA to understand this concretely. It has a built-in function Zeta[z] that you should compare with Sum[n^(-z), {n,1,100}] and Sum[(-1)^(n+1)*n^(-z), {n,1,100}]/(1-2^(1-z)) when z=2+14*I or z=1/2+14*I.

Once this is clear, you can look at the Taylor series $$F(z) = (z-1)\zeta(z) = \sum_{k=0}^\infty \frac{F^{(k)}(z_0)}{k!} (z-z_0)^k$$ which converges for every $z,z_0 \in \mathbb{C}$ (because $(z-1)\zeta(z)$ is an entire function)

Maclaurin Series of the Riemann Zeta

2 Answers2