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The trivial zeros of the Riemann zeta function are negative even integers. But I don't understand how that makes sense with the original definition of the function. $\zeta(-2) = \sum_{n=1}^{\infty} n^2$,

which is the sum of squares of natural numbers. How does this sum to zero? Am I missing something obvious?

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    $\begingroup$ The series is not how the function is defined for complex numbers $z$ with $\mathfrak R(z)\le 1$. $\endgroup$ – Regret Mar 15 '15 at 4:53
  • $\begingroup$ Can you elaborate on that? How do you define the function if the series diverges? $\endgroup$ – biryani Mar 15 '15 at 4:57
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    $\begingroup$ It is the analytic continuation of the original series definition, but to the complex plane. $\endgroup$ – Aloizio Macedo Mar 15 '15 at 4:59
  • $\begingroup$ I know very little about it, but this page appears to include a full definition of the function. $\endgroup$ – Regret Mar 15 '15 at 4:59
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    $\begingroup$ $\displaystyle \frac{1}{1-x}$ is the analytical continuation of $\sum_{n=0}^{\infty} x^n$ in the same way that the zeta is the analytical continuation of $\sum_{n=0}^{\infty} n^{-x}$ $\endgroup$ – Aloizio Macedo Mar 15 '15 at 5:03
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In short, the Riemann zeta function is not defined through the series you mention for its entire domain. Only for those complex $z$ with $\mathfrak R(z)>1$ is the Riemann zeta function defined by

$$\zeta(z)=\sum_{n=1}^\infty\frac1{n^z}$$

For all other complex numbers in its domain, an analytic continuation is used. That is why $\zeta(-2)=0$ can be true even though the series diverges. $\mathfrak R(-2)=-2\le1$, and as such, $\zeta(-2)$ is not defined through the series.

You can find the full definition of the Riemann zeta function here.

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Let us apply this answer, about the analytic extension of $\zeta$, for $m\in\mathbb{Z}$, $m\gt0$. Euler-Maclaurin Summation yields $$ \sum_{k=0}^nk^m=\zeta(-m)+\frac1{m+1}k^{m+1}+\sum_{k=1}^{m+1}a_kD^{k-1}x^m\tag{1} $$ where, as is described in this answer, $$ \begin{align} \sum_{k=0}^\infty a_kx^k &=\frac{x}{1-e^{-x}}\\ &=\frac x2\left(1+\coth\left(\frac x2\right)\right)\tag{2} \end{align} $$ Since $\frac x2\coth\left(\frac x2\right)$ is even, other than $a_1=\frac12$, $a_k=0$ for odd $k$.

Since the constant term in $(1)$ is $0$, and $D^mx^m=m!$, $$ \zeta(-m)+m!\,a_{m+1}=0\tag{3} $$ If $m\gt0$ is even, $a_{m+1}=0$, and therefore, $\zeta(-m)=0$ for positive, even $m$.

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You can see it by functional equation of Riemann zeta $$\zeta\left(s\right)=2^{s}\pi^{1-s}\sin\left(\frac{s\pi}{2}\right)\Gamma\left(1-s\right)\zeta\left(1-s\right).$$ We know that zeta has a pole only on $s=1$ and Gamma has poles in $0,-1,-2,...$. Some of these poles (for $s$ a negative odd number) are deleted by sine's zeros but the other? The only possibility is that zeta has zeros in every even negative integer.

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