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I found a derivation of $\zeta(-1) = -1/12$ that what I find nice because it follows the same steps as in those videos trying to show $1+2+3+\ldots = -1/12$, but making them rigorous :

$$\begin{eqnarray} \frac{2\eta(s)-1}{s} &=& \frac{1}{s}\sum_{n=1}^\infty (-1)^{n+1} (n^{-s}-(n+1)^{-s}) = \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1} x^{-s-1}dx\\ \frac{2\eta(s-1)-\eta(s)}{s} &=& \frac{1}{s}\sum_{n=1}^\infty (-1)^{n+1} n^{1-s}+ \frac{1}{s}\sum_{n=1}^\infty (-1)^{n+1} (n-1) n^{-s} \\ &=& \frac{1}{s}\sum_{n=1}^\infty (-1)^{n+1} n^{1-s}+ \frac{1}{s}\sum_{n=1}^\infty (-1)^{n} n (n+1)^{-s} \\ &=& \frac{1}{s} \sum_{n=1}^\infty (-1)^{n+1} n (n^{-s}-(n+1)^{-s}) = \sum_{n=1}^\infty (-1)^{n+1} n\int_n^{n+1} x^{-s-1}dx\end{eqnarray}$$ Both $\sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1} x^{-s-1}dx$ and $\sum_{n=1}^\infty (-1)^{n+1} n\int_n^{n+1} x^{-s-1}dx$ are Leibniz series valid (by analytic continuation) for $s > 0$, and their coefficients are bounded as $s \to 0$, so those two series stay bounded as $s\to 0$, therefore $2\eta(0)-1 = 0, \ 2\eta(-1)-\eta(0) = 0$ and

$$\eta(0) = 1/2, \qquad\eta(-1) =1/4$$

Using $\zeta(s) = \frac{\eta(s)}{1-2^{1-s}}$ we get $\zeta(-1) = -1/12 $.

Question : Can you generalize it to $\eta(-k), k \ge 2$ ? Do you know other elementary ways for obtaining $\zeta(-1) = -1/12$ ? $\scriptstyle\text{(elementary except for the analytic continuation part)}$

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closed as unclear what you're asking by Batominovski, Daniel W. Farlow, JonMark Perry, Chill2Macht, Alex Provost Aug 4 '16 at 21:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I don't really understand this question. Nevertheless, this link to Terence Tao's blog may be of interest, so I'll leave it here. $\endgroup$ – Will R Aug 4 '16 at 3:24
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    $\begingroup$ For $n>0$ and $n\in\mathbb N$, $$\zeta(1-2n) = \frac{(-1)^{n+1}B_{2n}}{2n(2\pi)^n}\sin\left(\frac{\pi(1-2n)}{2}\right)$$ where $B_k$ is the $k$th Bernoulli number. The second case is $$\zeta(-2n)=0$$ $\endgroup$ – Simply Beautiful Art Aug 19 '16 at 23:59
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    $\begingroup$ I also recommend not getting too hooked up on those math videos you find on the internet. They aren't really meant to be rigorous, just inspiring for the common folks. $\endgroup$ – Simply Beautiful Art Aug 20 '16 at 0:02
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An Elementary Non-Proof

Note that $\dfrac{1}{(1-z)^2}=\sum\limits_{k=0}^\infty\,(k+1)\,z^{k}$ leads to $$T:= 1-2+3-4+\ldots=\frac{1}{\big(1-(-1)\big)^2}=\frac{1}{4}\,.$$ Hence, if $S:=1+2+3+\ldots$, then $$S-T=4+8+12+\ldots=4\,(1+2+3+\ldots)=4\,S\,.$$ Thus, $$\zeta(-1)=S=-\frac{T}{3}=-\frac{1}{12}\,.$$

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    $\begingroup$ yes it is mentioned in wiki/1 + 2 + 3 + 4 + ⋯ but how do you justify it ? do you consider the analytic continuation of a function of two complex variables ? $\endgroup$ – reuns Aug 3 '16 at 22:31
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    $\begingroup$ As stated, I offered a non-proof. It makes no sense to try to justify something that is obviously wrong (due to divergence of both $S$ and $T$). I don't know what you actually mean by "elementary proofs" because I don't consider analytic continuation or regularization elementary. If you want something that can be justified but not elementary, user357980 offered that. My answer is not justifiable, but elementary. $\endgroup$ – Batominovski Aug 3 '16 at 22:38
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    $\begingroup$ I want to leave it as it is to illustrate how unclear questions can lead to answers like this. Stop acting like a dictator. $\endgroup$ – Batominovski Aug 20 '16 at 6:13
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    $\begingroup$ my question is not unclear for people knowing what is the definition of $\zeta(s)$ and $\eta(s)$, see the tag riemann-zeta $\endgroup$ – reuns Aug 20 '16 at 6:15
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    $\begingroup$ You have extremely thin skin, despite all your behavior. $\endgroup$ – Batominovski Aug 20 '16 at 6:19
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I don't know if this counts as elementary, but one of the various functional equations for $\zeta$(s): $$\zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)$$ may be helpful.

Evaluating this at $s = -1$ using the fact that $\Gamma(n+1) = n!$ and $\zeta(2) = \frac{\pi^2}{6}$, we get $$\zeta(-1) = -\frac{1}{2\pi^2}\sin\left(\frac{\pi}{2}\right)\Gamma(2)\zeta(2) = -\frac{1}{12}.$$ This generalizes, though the zeta function of positive odd numbers are not known in closed form. :(

The function $\xi(s)$ is used to define $\zeta(s)$ and satisfies $\xi(s) = \xi(1-s)$. I believe that this identity is a consequence of this and the reflection formula for $\Gamma$ given in the citations. (Also, this method seems related to yours given the functions involved.) Also, evaluating the second last equation in the second source also gives the value for $\zeta$.

However, I would add that both our methods are deeply related to the section "Zeta Function Regularization" in the Wiki article you added, though yours looks more like that, they are conceptually alike.

Cite: https://en.wikipedia.org/wiki/Riemann_zeta_function#The_functional_equation https://en.wikipedia.org/wiki/Reflection_formula

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  • $\begingroup$ it is not elementary at all $\endgroup$ – reuns Aug 3 '16 at 22:31

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