Riemann zeta function at $\Re(s)=0$ and $\Re(s)=1$

Question

Riemann zeta function at $\Re(s)=0$ and $\Re(s)=1$.

What happens to the zeta function at these points? For example $\sum_{n=1}^\infty \frac1{n^s}$ is defined for $\Re(s)>1$ and for $\Re(s)>0$ you have a different formula. But none of these include 0 or 1? Or does $\Re(S)>0$ include the 1? (or maybe it was defined wrong and should be $\Re(s)>0$ excluding $s=1$ in the book)

Answer 1

$\qquad\qquad\qquad\qquad\qquad\qquad\quad\zeta\big(1^\pm\big)=\pm\infty~$ and $~\zeta(0)=-\dfrac12$

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$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(s)$ for $\Re(s)=0.$

$\qquad\qquad\qquad\qquad$ The real part is in blue, and the imaginary part is in red.

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$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(s)$ for $\Re(s)=1.$

$\qquad\qquad\qquad\qquad$ The real part is in blue, and the imaginary part is in red.

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$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(s)$ for $\Re(s)=\dfrac12$

$\qquad\qquad\qquad\qquad$ The real part is in blue, and the imaginary part is in red.

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$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(x)$ for $x\in(-14,-1).$

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$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ The plot of $\zeta(x)$ for $x\in(-20,~1).$

$\quad$ Notice how each new “hump” to the left gets exponentially bigger than the previous one.

Answer 2

The zeta function, written in the series form you wrote down above, is only defined for the $Re(s)>1$, but it has an "analytic continuation" that is defined everywhere on the complex plane (not just $Re(s)>0$) except at $s=1$ where it has a simple pole. Thus it will have a well defined value at $=0$, which is $\zeta(0)= -1/2$.

For more information I recommend the book by H.M. Edwards, "Riemann's Zeta Function". It might also be useful to look up the notion of analytic continuation in any complex analysis textbook.