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According to the Theorem 12.7 of the book Analytic Nymber Theory by Apostol, $$\zeta(1-s) = 2(2\pi)^{-s} \Gamma(s) \cos \big(\frac{\pi s}{2}) \zeta(s)$$ which results in (as the book also says) that $\zeta(-2n) =0$ for $n=1,2,3, \dots$, the so-called trival zeros of $\zeta(s)$.

But how on earth $\zeta(-2n) = \sum_{i=1}^{\infty} \frac{1}{i^{-2n}}= \sum_{i=1}^{\infty} i^{2n}=\infty=0$?

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    $\zeta(-2n)$ is the analytic continuation of the the sum you mentioned (beyond $\Re(n)<-1/2$) and NOT the sum itself. This continuation is achived for example by the functional equation. That's the same idea that makes $\sum_{n\geq1}n"="\zeta(-1)=-\frac{1}{12}$ – 2017-02-09

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In the same spirit, we have $$2^0+2^1+2^2+2^3+\ldots =-1.$$ The seeming paradox is that the sum on the left is defined as the analytic continuation of the series $\sum_{k=0}^{\infty}z^k=\frac{1}{1-z}$ outside its original domain of convergence $|z|<1$.

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    After the edit, I find this answer quite unsatisfactory. The sum on the left is equal to $\infty$, at least _for the standard meaning of "sum"_. The point is that for it to be equal $-1$ we have to introduce a different notion of "sum". – 2017-02-09
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    @user1952009 My first comment refered to an old version of this answer. I forgot to delete it. – 2017-02-09
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Notice, that the Riemann zeta function is only defined when the $\Re\left(\text{s}\right)>1$, so:

$$\zeta\left(\text{s}\right)=\sum_{\text{n}\in\mathbb{N}^+}\frac{1}{\text{n}^\text{s}}\space\space\space\to\space\space\space\zeta\left(-2\text{s}\right)\ne\sum_{\text{n}\in\mathbb{N}^+}\frac{1}{\text{n}^{-2\text{s}}}=\sum_{\text{n}\in\mathbb{N}^+}\text{n}^{2\text{s}}\tag1$$

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    the comment above says $\Re(n)<-1/2$ ... – 2017-02-09
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    @Edi $\zeta(n)$ sum converges if $Re(n)>1$. The sum for $\zeta(-2n)$ converges for $Re(n)<-1/2$. – 2017-02-09
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    Slight correction: $\zeta(s)$ is only defined _by the sum_ when $Re(s)>1$. Everywhere else it is defined by different means. – 2017-02-09