# Trivial zeros of the Riemann Zeta function

A question that has been puzzling me for quite some time now:

Why is the value of the Riemann Zeta function equal to $0$ for every even negative number?

I assume that even negative refers to the real part of the number, while its imaginary part is $0$.

So consider $-2$ for example:

$f(-2) = \sum_{n=1}^{\infty}\frac{1}{n^{-2}} = \frac{1}{1^{-2}}+\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\dots = 1^2+2^2+3^2+\dots = \infty$

What am I missing here?

• The series def. is only valid for $\Re \{s\}>1$
– jcg
Mar 25 '14 at 18:43
• The Riemann-Zeta function is not the sum $\sum_{n=1}^\infty\frac{1}{n^s}$. It is the analytic continuation of this sum. Mar 25 '14 at 18:45
• The series definition of $\zeta(s)$ is valid only for $\Re(s)>1$. The zeros of $\zeta(s)$ at the negative even integers can be seen from looking at the functional equation. Nov 23 '16 at 17:45
• You may also interpret $\zeta(-2n)$ as an sum in the sense of Abel: $$\zeta(-2 k)\equiv\lim_{z \rightarrow 1_-}\sum_{n\geq 1}n^{2k}z^k$$ Nov 24 '16 at 3:41
• I wanted to know the answer to this for some time as well, but never got around to asking. +1 Nov 24 '16 at 6:26

For $$\mathrm{Re}(s)\gt1$$, $$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}\tag{1}$$ For $$\mathrm{Re}(s)\le1$$, we need to use Analytic Continuation and another formula for $$\zeta(s)$$ since $$(1)$$ does not converge for $$\mathrm{Re}(s)\le1$$.

One formula that can be used is the Functional Equation for $$\zeta$$ which says $$\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}=\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}\tag{2}$$ The symmetric form in $$(2)$$ is given as $$(14)$$ on MathWorld.

Since the recurrence for $$\Gamma$$ says that for $$n\in\mathbb{Z}$$ and $$n\le0$$, $$\frac1{\Gamma(n)}=0$$, we get that $$\zeta(2n)=0$$ for $$n\in\mathbb{Z}$$ and $$n\lt0$$.

Analytic Continuation of $$\boldsymbol{\zeta}$$ Using $$\boldsymbol{\eta}$$ and Integration by Parts

Define $$\eta$$, the alternating $$\zeta$$ function, as \begin{align} \eta(s) &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\\ &=\sum_{n=1}^\infty\frac1{n^s}-2\sum_{n=1}^\infty\frac1{(2n)^s}\\[6pt] &=\zeta(s)\left(1-2^{1-s}\right)\tag{3} \end{align} Formula $$(3)$$ follows since the terms of an alternating series are the terms of the non-alternating series minus twice the even terms.

Formula $$(3)$$ increases the domain to $$\operatorname{Re}(s)\gt0$$. $$\eta$$ also comes in handy to define $$\eta(s)\,\Gamma(s)=\int_0^\infty\frac{t^{s-1}}{e^t+1}\,\mathrm{d}t\tag{4}$$ which also converges for $$\operatorname{Re}(s)\gt0$$. However, we can integrate $$(4)$$ by parts $$k$$ times to get $$\bbox[5px,border:2px solid #C0A000]{\eta(s)\,\Gamma(s)=\frac{(-1)^k}{s(s+1)\cdots(s+k-1)}\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\frac1{e^t+1}\,\mathrm{d}t}\tag{5}$$ $$(5)$$ agrees with $$(4)$$ for $$\operatorname{Re}(s)\gt0$$ and converges for $$\operatorname{Re}(s)\gt-k$$. Thus, $$(5)$$ gives an analytic continuation of $$\zeta(s)$$ for $$\operatorname{Re}(s)\gt-k$$.

Using this method, $$\zeta(0)=-\frac12$$ is computed in this answer and $$\zeta(-1)=-\frac1{12}$$ is computed in this answer.

Proving $$\boldsymbol{\zeta(-2n)=0}$$

Note that $$(5)$$ can be rewritten as $$\eta(s)\,\Gamma(s+k)=(-1)^k\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\tag{6}$$ Setting $$s=1-k$$ in $$(6)$$ gives \begin{align} \eta(1-k) &=(-1)^k\int_0^\infty\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\\[6pt] &=(-1)^{k-1}\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{7} \end{align} Set $$k=2n+1$$ and we get $$\eta(-2n) =\frac{\mathrm{d}^{2n}}{\mathrm{d}t^{2n}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{8}$$ For $$n\ge1$$, the right side of $$(8)$$ is an odd function, so evaluating at $$t=0$$, and applying $$(3)$$, yields $$\bbox[5px,border:2px solid #C0A000]{\zeta(-2n)=0}\tag{9}$$

Analytic Continuation of $$\boldsymbol{\zeta}$$ Using the Euler-Maclaurin Sum Formula

As described in this answer, this answer, and this answer, for $$\operatorname{Re}(s)\gt-2m-1$$, $$\zeta(s)$$ can be represented as $$\hspace{-12pt}\bbox[5px,border:2px solid #C0A000]{\zeta(s)=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac{1}{k^s}-\left(\frac{1}{1-s}n^{1-s}+\frac12n^{-s}-\sum_{k=1}^m\frac{B_{2k}}{2k}\binom{s+2k-2}{2k-1}n^{-s-2k+1}\right)\right]}\tag{10}$$ where the formula in the parentheses is obtained from the Euler-Maclaurin Sum Formula.

Proving $$\boldsymbol{\zeta(-2m)=0}$$

If we plug $$s=-2m$$ into $$(10)$$, for $$m\ge1$$, the formula in parentheses exactly matches the sum outside the parentheses by Faulhaber's Formula. In fact, the Euler-Maclaurin Sum Formula is one way to prove Faulhaber's Formula. This means that $$\bbox[5px,border:2px solid #C0A000]{\zeta(-2m)=0}\tag{11}$$ We can also plug $$s=-2m+1$$, for $$m\ge1$$, into $$(10)$$ and note that Faulhaber's Formula does not include the constant term. Thus, we get $$\zeta(-2m+1)=-\frac{B_{2m}}{2m}\tag{12}$$

Functional Equation for $$\boldsymbol{\zeta}$$

The Fourier Transform of $$e^{-\pi x^2t}$$ is \begin{align} \int_{-\infty}^\infty e^{-\pi x^2t}e^{-2\pi ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-\pi(x-i\xi/t)^2t}e^{-\pi\xi^2/t}\,\mathrm{d}x\\ &=\frac1{\sqrt{t}}e^{-\pi\xi^2/t}\tag{13} \end{align} Applying the Poisson Summation Formula to $$(13)$$ says that $$1+2\sum_{n=1}^\infty e^{-\pi n^2t} =\frac1{\sqrt{t}}+\frac2{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\tag{14}$$ Note that for $$s\gt1$$, \begin{align} &\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}\\ &=\sum_{n=1}^\infty\frac1{n^s}\int_0^\infty e^{-\pi t}t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=\int_0^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=\int_0^1\left(\frac1{2\sqrt{t}}-\frac12+\frac1{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t +\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{1-\large s}2}\,\frac{\mathrm{d}t}t +\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)\left(t^{\frac{1-\large s}2}+t^{\frac{\large s}2}\right)\,\frac{\mathrm{d}t}t\tag{15} \end{align} The following integral is increasing in $$\alpha$$. For $$\alpha\ge0$$, we have \begin{align} \int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^\alpha\,\mathrm{d}t &=\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\int_{\pi n^2}^\infty e^{-t}\,t^\alpha\,\mathrm{d}t\\ &\le\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\left(\int_{\pi n^2}^\infty e^{-t}\,\mathrm{d}t\right)^{1/2}\left(\int_{\pi n^2}^\infty e^{-t}t^{2\alpha}\,\mathrm{d}t\right)^{1/2}\\ &\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac{e^{-\pi n^2/2}}{n^{2+2\alpha}}\\ &\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac2{\pi n^{4+2\alpha}}\\ &=\frac{2\,\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+2}}\zeta(4+2\alpha)\tag{16} \end{align} Thus, the last integral in $$(15)$$ defines an entire function. Therefore $$(15)$$ defines $$\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}$$ for all $$s\in\mathbb{C}$$. Since $$(15)$$ is invariant under $$s\leftrightarrow1-s$$, we have $$\bbox[5px,border:2px solid #C0A000]{\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}} =\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}}\tag{17}$$

• Would the downvoter care to comment? I've just spent a lot of time expanding my answer and would really like to know if there is something wrong.
– robjohn
Nov 24 '16 at 6:40
• I agree, its a wonderful answer. Nov 24 '16 at 12:41
• I have improved and reincorporated the section about the Euler-Maclaurin Sum Formula.
– robjohn
Nov 25 '16 at 15:16
• You've done enough, and the answer is wonderful as is. No need to derive more than all the formulas on the wikipedia page XD Nov 25 '16 at 23:53
• Rob, this is an excellent composition. +1 only, but wish I could give more. Nov 29 '16 at 20:27

The zeta function is defined as a meromorphic function given by

$$\zeta(s):=1^{-s}+2^{-s}+3^{-s}+\dots=\sum_{n=1}^\infty\frac1{n^s}\ \forall\ \Re(s)>1$$

Though it is common misconception to apply this definition whenever $\Re(s)\le1$, since clearly,

$$1+2+3+\dots$$

doesn't make much sense, normally. To define things like $\zeta(0)$, we use something called analytic continuation (very important part), which allows us to make sense of things. For example,

$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}\ \forall\ \Re(s)>0\tag!$$

which is not only true for $\Re(s)>1$ but also defined for $\Re(s)>0$. Here, we find that

$$\zeta(0)=\lim_{s\to0^+}\frac1{1-2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=-\frac12$$

We also have the nice reflection formula

$$\zeta(-s)=2^{-s}\pi^{-(s+1)}\sin\left(\frac{-s\pi}2\right)\Gamma(s+1)\zeta(s+1)$$

which can be used to show that

$$\zeta(-2k)=0\ \forall\ k\in\{1,2,3,\dots\}$$

As a short little proof of $(!)$ above, see that if we have

$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}\tag{converges for \Re(s)>0}$$

then, by subtracting out the odd terms and adding in the even terms, we are left with double the even terms:

$$\zeta(s)-\eta(s)=\sum_{n=1}^\infty\frac2{(2n)^s}=2^{1-s}\sum_{n=1}^\infty\frac1{n^s}=2^{1-s}\zeta(s)$$

$$\zeta(s)-\eta(s)=2^{1-s}\zeta(s)$$

$$(1-2^{1-s})\zeta(s)=\eta(s)$$

$$\zeta(s)=\frac1{1-s^{1-s}}\eta(s)$$

And that is how one can expand the domain of the zeta function.

Remark: Deriving other representations of the zeta function are usually much more hairy than above and often times require more areas of mathematics than you might first think. However, if you are so inclined, taking the Taylor expansion of $f(s)=s\zeta(s+1)$ should provide a series form of the zeta function converging almost everywhere.

• I added the details below. Yes it is not trivial, but it is not much more complicated than what you wrote (and it is easier than proving the functional equation) Nov 24 '16 at 12:49
• @user1952009 Yeah, I noticed (and +1), but anyone who easily understands your answer probably doesn't need to be here to learn very much. I'd likely bet most of the traffic here are those who want to read an answer like mine and to understand it without so much experience. (and yes, the functional equation above is far less trivial, but my presentation is simply less overbearing.) Nov 24 '16 at 12:55
• Sure my answer isn't addressed to the OP, and I wrote it because of the tag 'riemann-zeta' see what google answers Nov 24 '16 at 12:59
• @user1952009 I also wrote a simple approach to deriving a form of the zeta function that can be used for $s\ne1$ in the last line. Nov 24 '16 at 13:08
• The more you understand $\zeta(s)$, the more you understand how we have no clue for proving the RH (because each of those formulas and theorems can be generalized to functions for which the RH fails) Nov 24 '16 at 13:42

The Zeta function is defined as $\zeta(s)=\sum_{n\ge1}n^{-s}$ only for $s\in\mathbb{C}$ with $\Re(s)>1$!

The function on the whole complex plane (except a few poles) is the analytic continuation of that function.

On the Wikipedia page, you can find the formula: $$\zeta(s)=\frac{2^{s-1}}{s-1}-2^s\int_0^\infty\frac{\sin(s\arctan t)}{(1+t^2)^{\frac{s}{2}}(e^{\pi t}+1)}dt$$ for $s\neq 1$. Maybe working on this integral for $s$ a negative integer will give you the result.

• Thank you @Daniel Robert-Nicoud. So from your answer, as well as the other answers and comments here, I sort of understand that the Riemann Zeta function is actually some sort of "extension" to the function specified in my question. Is that correct? Mar 25 '14 at 18:49
• @barakmanos Yes, exactly. The idea is that you can extend the function (defined only in some subset of the complex plane, namely you want that the real part is greater than $1$) to a meromorphic function (uniquely determined, by the identity theorem). The function I posted is said extension (even though I have no idea how it can be found). Mar 25 '14 at 18:51
• So what exactly is the relation between the function specified in your answer and the function specified in my question? Mar 25 '14 at 18:52
• Sorry, rolled-back your edit by mistake; Can you redo it? Mar 25 '14 at 18:56
• It's easier to determine the values at the negative integers using the contour integral representation. Mar 25 '14 at 19:11

There is a way to prove that $\zeta(-2K) = 0$ :

• By definition of the Bernouilli numbers $\frac{z}{e^z-1}= \sum_{k=0}^\infty \frac{B_k}{k!}z^k$ is analytic on $|z| < 2\pi$

• Note that $\frac{z}{e^z-1}-\frac{z}{2} = \frac{z}{2}\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}$ is an even function, therefore $\frac{z}{e^z-1}-1-\frac{z}{2}=\sum_{k=2}^\infty \frac{B_k}{k!}z^k$ is an even function, and $B_{2k+1}=0$ for $k\ge 1$.

• For $Re(s) > 0$, let $\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}dx$. It converges absolutely so it is analytic.

Integrating by parts $\Gamma(s+1) = s \Gamma(s)$ providing the analytic continuation to $Re(s) \le 0$ : $\Gamma(s) = \frac{\Gamma(s+k)}{\prod_{m=0}^{k-1} s+m}$.

Thus $\Gamma(s)$ is analytic on $\mathbb{C} \setminus -\mathbb{N}$ with poles at the negative integers where $\Gamma(s) \sim \frac{(-1)^k}{k!}\frac{1}{s+k}$

• With the change of variable $x = ny$ you have $\Gamma(s) n^{-s} = \int_0^\infty x^{s-1} e^{-nx}dx$ so that for $Re(s) > 1$ where everything converges absolutely $$\Gamma(s) \zeta(s) = \sum_{n=1}^\infty \int_0^\infty x^{s-1} e^{-nx}dx= \int_0^\infty x^{s-1}\sum_{n=1}^\infty e^{-nx}dx=\int_0^\infty x^{s-2}\frac{x}{e^x-1}dx$$

• Note that $\frac{1}{s+k-1} = \int_0^1 x^{s-2+k}dx = \int_0^\infty x^{s-2} x^{k}1_{x < 1}dx$ so that $$\Gamma(s) \zeta(s)- \sum_{k=0}^{K}\frac{B_k}{k!}\frac{1}{s+k-1} =\int_0^\infty x^{s-2}\left(\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!}x^k1_{x < 1}\right)dx \tag{1}$$ Now as $x \to 0$ :$\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!}x^k\sim \frac{B_{K+1}}{(K+1)!}x^{K+1}$ and hence $(1)$ converges and is bounded for $Re(s) > -K$, i.e. as $s \to -k$ : $$\frac{(-1)^k}{k!}\frac{1}{s+k} \zeta(s) \sim\Gamma(s) \zeta(s)\sim \frac{B_{k+1}}{(k+1)!}\frac{1}{s+k}$$ whence $$\boxed{\zeta(-k) = (-1)^k\frac{B_{k+1}}{k+1} \implies \zeta(-2k) = 0, k \in \mathbb{N}^*}$$

• Since the Bernoulli numbers are the coefficients for the Euler-Maclaurin Sum Formula, this agrees with the recent addition to my answer. (+1)
– robjohn
Nov 24 '16 at 7:04

As others have pointed out, that's not quite the definition of the zeta function. The zeta function is in fact the unique meromorphic function that's equal to that wherever that exists. (To prove uniqueness, you can use Taylor series and the theorem that such a function is equal on any disc where it exists to the Taylor series at the center.)

Regarding the given value, I'll give Riemann's original proof:

$$\Gamma(x) = \int_0^\infty\frac{t^{x-1}}{e^t}dt = \int_0^\infty\frac{(nt)^{x-1}}{e^{nt}}ndt = n^x\int_0^\infty \frac{t^{x-1}}{e^{nt}}dt$$

Therefore

$$\Gamma(x)\zeta(x) = \sum_{n=1}^\infty\frac{\Gamma(x)}{n^x} = \sum_{n=1}^\infty\int_0^\infty \frac{t^{x-1}}{e^{nt}}dt = \int_0^\infty t^{x-1}\sum_{n=1}^\infty\frac{1}{(e^{t})^n}dt = \int_0^\infty\frac{t^{x-1}}{e^{t} - 1}dt$$

Now for the slightly tricky part: this still isn't defined for $\Re[x] < 1$. To get around this, what he did was use the identity $2i\sin(\pi x) = e^{i\pi x}-e^{-i\pi x}$ to get

$$2i\sin(\pi x)\Gamma(x)\zeta(x) = (e^{i\pi x}-e^{-i\pi x})\int_0^\infty\frac{t^{x-1}}{e^{t} - 1}dt = \int_0^\infty\frac{(e^{-i\pi}t)^{x-1}}{e^{t} - 1}dt - \int_0^\infty\frac{(e^{i\pi}t)^{x-1}}{e^{t} - 1}dt$$

(Note I've left $e^{\pm i\pi}$ unsimplified, since you'll get different values for each when x isn't an integer.) What Riemann did was he treated this as a contour integral, replacing the hairpin turn at the origin with a tight loop - which doesn't change the value since the integrand is analytic everywhere in the origin's neighborhood. So what you ultimately have is

$$\zeta(x) = \lim_{s\to x}\frac{\int_C\frac{(-t)^{s-1}}{e^t - 1}dt}{2i\sin(\pi s)\Gamma(s)}$$

Where the contour treats -1 as $e^{i\pi}$ on the way in, $e^{-i\pi}$ on the way out, and goes around the origin in a tight loop in a clockwise (i.e., negative) direction. Fortunately, in the case of integers positive or negative, there's no branch cut, so the way in and the way out cancel, leaving only a tight loop that can be computed with the residue theorem. So with that in mind,

$$\int_C\frac{(-t)^{s-1}}{e^t - 1}dt = -\int_C (-t)^{s-2}\frac{t}{e^t - 1}dt$$

It turns out that the coefficients of the Maclaurin expansion of $\frac{x}{e^x-1}$ are just the Bernoulli numbers divided by the factorials, and every other Bernoulli number starting with the fourth is zero - meaning that the coefficient of every odd exponent of the Maclaurin series but $x^1$ is zero, so where s is a negative even integer, the residue is zero, so the integral is zero.

So there's still one hurdle - the gamma function has poles at nonpositive integers, and $\sin(\pi x)$ is equal to zero. That's easy enough to work out, though, using the property $s\Gamma(s) = \Gamma(s+1)$ (which if you didn't know already, integration by parts will show you is implied by the definition above), and the fact that x is a negative even integer,

$$\lim_{s\to x}\frac{\frac{1}{\Gamma(s)}}{\sin(\pi s)} = \lim_{s\to x}\left(\frac{\prod_{n=0}^{-x}(s-x-n)}{(s-x)\Gamma(s-x+1)}\cdot\frac{(s-x)}{\sin(\pi s)}\right) = \frac{(-x)!}{\pi}$$

What this is isn't important, only that now you have zero in the numerator and a nonzero denominator, so you have zero.

• Oh man, there comes Ramanujan in with contour integrals for my real integral problems. Hefty little answer for me to go through, but much enjoyable. +1 Nov 24 '16 at 2:55
• @SimpleArt you mean Riemann, right? Nov 25 '16 at 5:24
• @BrevanEllefsen Oh, yeah. XD Its ok, the both start with the letter "R" and end with an "n", so its close enough. Nov 25 '16 at 12:39

Here is some nice info: https://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation

To answer your question, there is a functional equation that the zeta function satisfies: $$\zeta(s)=2^s\pi^{s-1}\sin\Big( \frac{\pi s}{2} \Big) \Gamma(1-s)\zeta(1-s),$$ where $\Gamma$ is the Gamma function. For negative even integers $k$ (-2, -4 et cetera), the sine factor is zero, which makes $\zeta(k)=0$.

• This doesn't actually the answer the question concerning the given series. Nov 23 '16 at 17:54
• True, although it gives a reason why the zeta function is zero at the negative even numbers. I suppose the other answers are better suited for an accept. Nov 23 '16 at 17:59

Other answers have addressed in detail the specifics of the zeta function. This is just to add a simple note which, I think, goes to the heart of the OP's question as to what he or she is missing, namely how can it make sense to talk about the behavior of a function defined by an infinite series at a value where the series diverges.

The simplest way to understand this is with the geometric series

$$f(x)=1+x+x^2+x^3+\cdots$$

This series clearly diverges if $$|x|\gt1$$. Yet it makes sense to say $$f(2)=-1$$, because of the identity

$$1+x+x^2+x^3+\cdots={1\over1-x}$$

which holds for all $$|x|\lt1$$. The right hand side is clearly meaningful for all $$x\not=1$$; it provides what's called an analytic continuation of the function $$f$$.

Something similar (but more complicated) occurs with the zeta function: The series definition $$\zeta(s)=\sum n^{-s}$$ makes no sense when the real part of $$s$$ is less than $$1$$, but there are identities with expressions that do. One of the things to look forward to, when you take a course in complex analysis, is learning why, for functions like the geometric series and the zeta function, you don't get conflicting identities that give different values for the analytic continuation, which is why it makes sense to talk about "the" zeta function $$\zeta(s)$$.