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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 $\zeta(-1) = -1/12$ ? $\scriptstyle\text{(elementary except for the analytic continuation part)}$

reuns
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2 Answers2

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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, $$\zeta(-1)=S=-\frac{T}{3}=-\frac{1}{12}\,.$$

Batominovski
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    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 ? – reuns Aug 03 '16 at 22:31
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    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. – Batominovski Aug 03 '16 at 22:38
  • @user1952009 If you want to put your new proof, please create your own answer (in your own question, for example). I prefer to leave my non-proof as it is. – Batominovski Aug 20 '16 at 06:12
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    I want to leave it as it is to illustrate how unclear questions can lead to answers like this. Stop acting like a dictator. – Batominovski Aug 20 '16 at 06:13
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    my question is not unclear for people knowing what is the definition of $\zeta(s)$ and $\eta(s)$, see the tag riemann-zeta – reuns Aug 20 '16 at 06:15
  • I know perfectly well what those functions are. Do you still want to bring back the argument why your question is unclear? I'm tired of this. – Batominovski Aug 20 '16 at 06:16
  • I insulted you? No, you started all this. And I will keep rolling back my answer. This is my answer, not yours. – Batominovski Aug 20 '16 at 06:17
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    You have extremely thin skin, despite all your behavior. – Batominovski Aug 20 '16 at 06:19
  • no, I'm just a math student interested in $\zeta(s)$, see all my questions/answers, and this question "ways to show $\zeta(s) = -1/12$ other than with the residue theorem and the functional equation" is interesting – reuns Aug 20 '16 at 06:20
  • You know what? There were four other people who thought your question was unclear and they voted to close it down. Nobody tried to reopen it. Let that speak for itself. – Batominovski Aug 20 '16 at 06:24
  • I see. It's everybody else's false. You are not in the wrong. You never insulted anybody. I'm done with you, at least for now. Do not change my answer. – Batominovski Aug 20 '16 at 06:28
  • how can I be wrong when proposing a new proof of $\zeta(-1) = -1/12$ on a math forum, and asking people if they know others that are of the same kind ? – reuns Aug 20 '16 at 06:29
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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 $$\zeta(-1) = \frac{1}{2\pi^2}\sin\left(-\frac{\pi}{2}\right)\Gamma(2)\zeta(2) = \frac{1}{2\pi^2} \cdot (-1) \cdot 1! \cdot \frac{\pi^2}{6} = -\frac{1}{12}.$$ 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

Ryu Dae Sick
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user357980
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