Maclaurin Series of the Riemann Zeta 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?
$\sum_{n=1}^{\infty} \frac{1}{n^0}$  - $\sum_{n=2}^{\infty} \frac{\ln(n) }{n^0 \cdot 1!}s$ + $\sum_{n=2}^{\infty} \frac{\ln(n)^2 }{n^0 \cdot 2!}s^2$ -  $\sum_{n=2}^{\infty} \frac{\ln(n)^3 }{n^0 \cdot 3!}s^3$ $+ ... +$  $(-1)^x\sum_{n=2}^{\infty} \frac{\ln(n)^x }{n^0 \cdot x!}s^x$
 A: 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,
$$a_n=\lim_{x\to-1^+}\sum_{j=0}^n\sum_{k=1}^\infty\frac{x^{k+1}\ln^j(k)}{j!}\sum_{l=0}^\infty\frac{(-l\ln(2))^{n-j}2^l}{(n-j)!}$$
A: 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)
