What is the analytic continuation of the Riemann zeta function?

There is only one "normal" definition of the Zeta function. For $\operatorname{Re}(s) > 1$, the zeta function is defined as $\displaystyle \sum_{k=1}^{\infty} \dfrac1{k^s}$. For the rest of the $s$ in the complex plane, it is defined as the analytic continuation of the above function. The functional equation $$\zeta(s) = 2^s \pi^{s-1} \sin \left(\dfrac{\pi s}2\right) \Gamma(1-s) \zeta(1-s)$$ can be used to obtain the value of the $\zeta$ function for $\operatorname{Re}(s) < 1$, using the value of the zeta function for $\operatorname{Re}(s)>1$.

If you evaluate the series $\displaystyle \sum_{k=1}^{\infty }\dfrac1{k^s}$ for $\text{Real}(s) \leq 1$, the series will not converge. So you cannot evaluate the series at any of the zeros, let alone non-trivial zeros.

We have functional equation $$ \zeta(s)=2^{s}\pi^{s-1}\sin\Bigl(\frac{\pi s}{2}\Bigr)\Gamma(1-s)\zeta(1-s). $$ We can prove this using contour integral.
Using this formula, we can expand Riemann Zeta Function to the whole complex plane except $s\neq1$.

To find the zeros of Riemann Zeta Function, you can use the well-known formula $$\left|Z(t)\right|=\left|\zeta(1/2+it)\right|$$ where $$Z(t)=\zeta\left(1/2+it\right)\frac{\Gamma(1/4+it/2)\pi^{-1/4-it/2}}{\left|\Gamma(1/4+it/2)\pi^{-1/4-it/2}\right|}$$ for real $t$.

If by "normal definition of the Riemann zeta-function" you mean $$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$$ well, the thing about that series is that it doesn't converge for real part of $s$ less than or equal to $1$. In particular, it doesn't converge at any of the zeros of the zeta-function, trivial or otherwise.

To understand how the zeta-function is defined for real part less than or equal to 1, you have to be familiar with "analytic continuation." Are you?

Let

$$\zeta(n) = \frac{1}{1^n} + \frac{1}{2^n} + \frac{1}{3^n} + \frac{1}{4^n} + ... $$

Then

$$\frac{\zeta(n)}{2^{n-1}} = \frac{1}{1^n2^{n-1}} + \frac{1}{2^n2^{n-1}} + \frac{1}{3^n2^{n-1}} + \frac{1}{4^n2^{n-1}} + ...=\frac{2}{2^n} + \frac{2}{4^n} + \frac{2}{6^n} + \frac{2}{8^n} + ...$$

Hence

$$\zeta(n)-\frac{\zeta(n)}{2^{n-1}}=\frac{1}{1^n} - \frac{1}{2^n} + \frac{1}{3^n} - \frac{1}{4^n} + ... $$

and so

$$\zeta(n) = \frac{1}{1-\frac{1}{2^{n-1}}} \left(\frac{1}{1^n} - \frac{1}{2^n} + \frac{1}{3^n} - \frac{1}{4^n} + ...\right)$$

which has a larger convergence range compared to the original definition.

To evaluate each term,

$$\frac{1}{k^{a+ib}} $$

$$= \frac{1}{k^a} \frac{1}{k^{ib}}=\frac{1}{k^a}\frac{1}{e^{i(b \log k)}}= \frac{e^{-i(b \log k)}}{k^a}$$

$$=\frac{\cos(b \log k)}{k^a} - i\frac{\sin(b \log k)}{k^a}$$

which computes $\zeta(s)$ to any arbitrary degree of precision.

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The closed circular integral

$$ \oint \frac{1}{\zeta(s)} ds \ne 0$$

$$\sum_{circle fragments} \frac{1}{\zeta(s)} \Delta s \ne 0$$

signals the presence of at least one zero within.

To compute the inverse of $\zeta(s)=a+ib$, observe that

$$\left(\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}\right) \left(a+ib\right) = \left(\frac{a^2 + b^2}{a^2+b^2} - i\frac{ab-ab}{a^2+b^2}\right) = 1$$

and thus

$$\frac{1}{\zeta(s)} = \frac{1}{a+ib} = \left(\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}\right) $$

The first 3 zeros of $\frac{1}{\zeta(s)}$ occur at

$$1/2 ± 14.135i$$ $$1/2 ± 21.022i$$ $$1/2 ± 25.011i$$

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Analytic continuation of $\sum_{n=1}^\infty$ can be performed as follows:

Let

$$k_s = \int_0^\infty \frac{y^{s-1}}{e^y} dy$$

Since the integral clearly converges for any value of $s$, then $k_s$ is a certain finite number defined for any given $s$.

Parametrising $y$ by $nx$,

$$k_s = \int_0^\infty \frac{(nx)^{s-1}}{e^{nx}} (n dx)$$

$$= n^s \int_0^\infty \frac{x^{s-1}}{e^{nx}} dx$$

Therefore

$$\frac{1}{n^s} = \frac{1}{k_s}\int_0^\infty \frac{x^{s-1}}{e^{nx}}dx$$

and

$$\sum_{n=1}^\infty \frac{1}{n^s} = \sum_{n=1}^\infty \left(\frac{1}{k_s}\int_0^\infty \frac{x^{s-1}}{e^{nx}}dx\right)$$

$$= \frac{1}{k_s} \int_0^\infty x^{s-1} \left( \sum_{n=1}^\infty \frac{1}{e^{nx}}\right) dx$$

$$= \frac{1}{k_s} \int_0^\infty x^{s-1} \left( \frac{1}{e^x - 1}\right) dx$$

$$= \frac{1}{k_s} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx$$

Since the integral $\int_0^\infty \frac{x^{s-1}}{e^x - 1} dx$ has a finite derivative over the entire complex plane and converges for all $s$ including $s \le 1$, by the uniqueness of analytic continuation theorem,

$$\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{k_s} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx$$

must be the uniquely correct analytic continuation of $\sum_{n=1}^\infty \frac{1}{n^s}$.

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Computationally

$$k_s = \int_{0}^\infty \frac{y^{s-1}}{e^y} dy $$

$$\approx \sum_{y=0}^\infty \frac{y^{s-1}}{e^y} \Delta y $$

$$ k_{a+ib} = \sum_{y=0}^\infty \frac{y^{a + ib -1}}{e^y} \Delta y $$

$$ = \sum_{y=0}^\infty \frac{y^{a -1}}{e^y}e^{i(b\log y)} \Delta y $$

$$ = \sum_{y=0}^\infty \frac{y^{a -1}}{e^y}\left[ \cos(b\log y) + i \sin(b\log y) \right] \Delta y $$

$$ = \sum_{y=0}^\infty \frac{y^{(a-1)}\cos(b\log y)}{e^y}\Delta y+ i \sum_{y=0}^\infty \frac{y^{(a-1)}\sin(b\log y)}{e^y}\Delta y$$

and

$$\sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{k_s} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx$$

$$\approx \frac{1}{k_s} \sum_{x=0}^\infty \frac{x^{s-1}}{e^x - 1} \Delta x$$

$$\sum_{n=1}^\infty \frac{1}{n^{a+ib}} = \frac{1}{k_{a+ib}} \left( \sum_{y=0}^\infty \frac{x^{(a-1)}\cos(b\log x)}{e^x - 1}\Delta y+ i \sum_{y=0}^\infty \frac{x^{(a-1)}\sin(b\log x)}{e^x - 1}\Delta y \right)$$

where the final complex valued division can be evaluated as

$$\sum_{n=1}^\infty \frac{1}{n^s} \approx \frac{a+ib}{c+id} = \frac{ac+bd}{c^2+d^2} + i \left(\frac{bc-ad}{c^2+d^2}\right)$$

and it can be calculated that e.g.

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} \approx 1.645 + 0i$$

$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90} \approx 1.082 + 0i$$