# elementary ways to show $\zeta(-1) = -1/12$ [closed]

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

• 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. Aug 4, 2016 at 3:24
• 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 $$\zeta(-2n)=0$$ Aug 19, 2016 at 23:59
• 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. Aug 20, 2016 at 0:02

## 2 Answers

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

• 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 ? Aug 3, 2016 at 22:31
• 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. Aug 3, 2016 at 22:38
• I want to leave it as it is to illustrate how unclear questions can lead to answers like this. Stop acting like a dictator. Aug 20, 2016 at 6:13
• my question is not unclear for people knowing what is the definition of $\zeta(s)$ and $\eta(s)$, see the tag riemann-zeta Aug 20, 2016 at 6:15
• You have extremely thin skin, despite all your behavior. Aug 20, 2016 at 6:19

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.

• it is not elementary at all Aug 3, 2016 at 22:31
• But i like it. Helpful for me. Dec 16, 2020 at 10:45