Proof of $\zeta (-2)$ simple math.

Question

Is there a way to proof that $\zeta (-2)=0$ the same way that the following "proof" is constructed?

I am talking about a similar proof to the (following) one of $\zeta(-1)$. It would be really nice if you could help me out with this.

Consider the following series: $$A=1-1+1-1+1-1+1-1+1-1+1-1+...$$ This series, sometimes called Grandi's series and is technically undefined. But using arithmatic and simple summation, we can show that it equals to 0.5:

$$A=1-1+1-1+1-1+1-1+1-1+1-1+1-1+...$$ Now Addiding another A to this we can pair up two terms so that they each cancel out: $$A+A=1-(1-1)+(1-1)-(1+1)+(1-1)-(1+1)+...$$ Which means that $A+A=1-0+0-0+0-0+...=1$, so that $2A=1$ and thus $A=\frac {1}{2}$

Now consider the series: $$B=1-2+3-4+5-6+7-8+9-10+...$$

A similar thing appears here, if you add B to itself and offset this sum by one; $$2B=1-(2-1)+(3-2)-(4-3)+(5-4)+(6-5)-(7-6)+...=1-1+1-1+1-1+1-1+1-...$$ Which means that $$2B=A$$ thus $$B=\frac {A}{2}=\frac{1}{4}$$

Now consider $\zeta(-1)$, which "equals" (not really equals, but let's asume it does): $S=1+2+3+4+5+6+7+8+...$

Now consider $S-B$: $$1-1+2+2+3-3+4+4+5-6+6+6+7-7+8+8+9-9+10+10+11-11+12+12+13-13+...$$ $$=0+4+0+8+0+12+0+16+0+20+0+24+0+...$$ $$=4*(1+2+3+4+5+6+...)$$ $$=4*B$$ This means that $S-B=4*S$, so that $-B=3*S$ and thus: $\frac{-B}{3}=S$

If $\frac{-B}{3}=S$, then $S=\frac{-1}{3 \times 4}= \frac {-1}{12}$

In the end, my question would be, if there is this kind of proof for $\zeta(-2)$, if there is one, would you be so kind to explain it to me? Thanks.

Answer 1

None of these manipulations are valid in terms of finding $\zeta(-2)$, since you don't use the summation formula for $\zeta(s)$ when $s$ has real part less than $1$. You even mention this yourself at one point. To break down the flaws in your argument:

This series, sometimes called Grandi's series and is technically undefined. But using arithmatic and simple summation, we can show that it equals to 0.5:

It is only equal to $1/2$ if we use Cesaro convergence. Under typical definitions of infinite summation and convergence of infinite sequences, it is undefined. There is not a "technically undefined" here, because it is undefined. If it could be given a value in a rigorously justified manner without Cesaro, it would be given that value - manipulating divergent infinite sums is not justified under typical scenarios.

Now consider $\zeta(-1)$, which "equals" (not really equals, but let's asume it does): $S=1+2+3+4+5+6+7+8+...$

It doesn't equal it, and we cannot assume it equals it. You cannot assume something false and expect to get anything sensible from it.

For $\zeta(-1)$, we calculate its value through Riemann's functional equation,

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

which holds for $s$ with real part less than $1$ (which includes $s=-1$). There is no summation here. $\zeta(-1)$ is not defined in the same way as $\zeta(s)$ is defined for $s$ with real part at least $1$, i.e. a summation. Such summations will diverge. Instead, we use analytic continuation to obtain this functional equation and thus be able to define $\zeta$ for all $s \in \Bbb C - \{1\}$. However, this comes at the consequence of not defining $\zeta$ in terms of a summation for the left half-plane.

Thus, $\zeta(-1) \ne \sum_{n=1}^\infty n$. In fact, that sum is divergent, and so it is senseless to assign any value to it under typical assumptions, since its partial sums do not converge at all.

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The easiest way to find the value of $\zeta(-2)$ would be through the functional equation. Plugging it in, we see

$$\zeta(-2) = 2^{-2} \cdot \pi^{-2-1} \sin(-2\pi/2) \Gamma(1-(-2)) \zeta(1-(-2))$$

Simplifying,

$$\zeta(-2) = \frac{\pi^3}{4} \sin(-\pi) \Gamma(3) \zeta(3)$$

$\sin(k \pi)$ is $0$ for all integers $k$, and thus $\zeta(-2) = 0$.

In fact, whenever $s$ is a negative, even integer (i.e. $s=-2,-4,-6,-8,\cdots$), it can be shown $\zeta(s) = 0$ in those cases as well, thanks to the sine term.

Answer 2

When denoting the series as $ A $, you are assuming that $ A $ does exist in $ \mathbb{R} $, that the series does converge to a real value, that $ A $ is a real number on which we can perform arithmetic operations such as addition or multiplication with a constant. But this is the point on which you are mistaken : The series doesn't converge to any real value, in other words, it doesn't make any sens to denote the infinite summation as $ A $, or else you are assuming the sequence $ \left(\sum\limits_{k=0}^{n}{\left(-1\right)^{k}}\right)_{n} $ does converge to $ A\in\mathbb{R} \cdot $

Another point to discuss : The definition of $ \zeta $ as an infinite series does make sens, only for a complex value that has a real part strictly greater than $ 1 \cdot $ If I you're speaking about negative numbers than forget about this definition of $ \zeta $, and try to look for other definitions which are extensions of $ \zeta $ to the complex plane.