How $\sum_{i=1}^{\infty} i^{2n}=0$?

Question

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

Answer 1

Notice, that the Riemann zeta function is only defined when the $\Re\left(\text{s}\right)>1$, so:

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

Answer 2

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 $|z|<1$.