Other way show that $\zeta(-2)=(-\frac{1}{12})\mod 2$?

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Lemma: We knew that for any integer $a$ : ${a}^{p}=a \mod p$.

Then $1^p=1\mod p ,2^p=2\mod p ,3^p=3\mod p , \ \dots,\ n^p=n\mod p $.

Just to sum each term by term RHS and LHS we will get the following:

$1^p+2^p+3^p+ \dots + n^p= (1+2+3+ \dots +n)\mod p $.

let $ n$ go to $\infty$ we get:

$$\sum_{n=1}^{\infty}n^p= \sum_{n=1}^{\infty}n \mod p \tag{I}$$

let $p=2$ then if i'm right, the RHS of (I) =$-\frac{1}{12}$ according to Euler's proof and the LHS is $\zeta(-2)$. Hence we get:

$\zeta(-2)=-\frac{1}{12} \mod 2 $.

Is there someone show me a little bit if my proof to show that :$\zeta(-2)=-\frac{1}{12} \mod 2 $ is true, or does this make no sense?

Thank you for any kind of help.

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17
On BEST ANSWER

If you buy the silly formulation: $1+2+3+\dots = -1/12$, you still can't say that $1+2+3+\dots\equiv 1^p+2^p+\dots$, because the finite proof of $a+b\equiv a^p+b^p$ doesn't apply to infinite sequences.

For example: $1+3+3^2+\dots = -\frac{1}{2}$ in "some sense." But:

$$1^2+3^2+3^4\dots =-\frac{1}{8}$$

Is $$-\frac{1}{2}\equiv -\frac{1}{8}\pmod {2}?$$

No, it is not, under any useful definition of that expression.

We know that if $a_i\equiv b_i\pmod p$ then for any $n$, $$a_1+a_2+\cdots+a_n\equiv b_1+b_2+\cdots b_n\pmod p,$$ but we can't conclude from those finite expressions that:

$$a_1+a_2+\cdots =b_1+b_2 +\cdots \pmod p$$

We prove the first by induction, but infinite sums cannot be reached by induction - you can't just wave you hands and say "In the limit, they must be congruent." By the same reasoning, in the limit, $1+2+3+\cdots$ must be positive.

Take the first step in the "proof" of $1+2+3+\cdots=-\frac1{12}$:

$$1-1+1-1+1\cdots = \frac{1}{2}$$

Multiply both sides by $2$, and we'd get:

$$2-2+2-2+\cdots = 1$$

Therefore $0\equiv 1\pmod 2$! Is that a useful way of looking at modular arithmetic?

Ultimately, you have to define stuff and prove stuff. Go back to your definitions.

The damage done by the nonsense $1+2+3+\cdots = -\frac{1}{12}$ is not that it is wrong. It is wrong, technically, but if you rewrite it as $\zeta(-1)=-\frac{1}{12}$, it is true.

The damage is precisely the complete lack of definitions and proofs of what manipulations are allowed to "prove" this sort of sum. It gives novices the idea that they can play will infinite series willy-nilly without even the vaguest caution. Anything involving the infinite in mathematics requires care.

10
On

There are several foggy steps...

  • you add infinitely many numbers,
  • you use $1+2+3+\dots=-\frac1{12}$, which I personally do kind of believe in, but it obviously cannot be rigolously valid,
  • What does it mean to be congruent to $-\frac1{12}$ at all modulo $2$? Are we talking about only integers, or not?
  • Note also that you are talking about the sum of sequence $1,0,1,0,1,0,1,0,1,\dots$ (either as $n$ or $n^2$) modulo $2$.

The rest of logical arguing seems correct.