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There already questions such as $1 + 1 + 1 +\cdots = -\frac{1}{2}$ and Why does $1+2+3+\cdots = -\frac{1}{12}$? which show how $\zeta(0)$ and $\zeta(-1)$ can be calculated.

What are some ways to evaluate the Riemann zeta function at any negative integer that appears to have a direct correlation to $\sum_{n=1}^\infty\frac1{n^s}$? That is to say, results that can be attained by manipulating this series somewhat directly. So not things such as the reflection formula or the Bernoulli numbers which don't seem to relate to the above series.

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  • $\begingroup$ Took a quick search but didn't find a dupe, but perhaps there is one I'm not aware of... $\endgroup$ – Simply Beautiful Art Sep 4 at 11:36
  • $\begingroup$ This page from the OEIS wiki gives the formula $\zeta(-s) = 2^{-s} \pi^{-s-1} \sin \left(-\frac{\pi s}{2} \right) \Gamma(1+s) \zeta(1+s)$ for $\Re(s) > 0$. $\endgroup$ – automaticallyGenerated Sep 4 at 13:09
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    $\begingroup$ @automaticallyGenerated, I am pretty sure that OP already know the methods using reflection formula and/or the Bernoulli numbers, as he/she is explicitly asking an answer which does not use them. $\endgroup$ – Sangchul Lee Sep 4 at 14:02
  • $\begingroup$ Oh I see. I should have read the question more carefully. $\endgroup$ – automaticallyGenerated Sep 4 at 14:33
  • $\begingroup$ In that case, try looking at this $\endgroup$ – automaticallyGenerated Sep 4 at 14:34
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One can relate the Riemann zeta function $\zeta(s)$ to the Dirichlet eta function $\eta(s)$ using

$$ (1-2^{1-s})\zeta(s) =(1-2^{1-s})\sum_{n=1}^\infty\frac{1}{n^s} =\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} =\eta(s)$$

which follows quite easily by multiplying out the LHS. This identity holds a priori for $s > 1$, and then extends to all of $s\in\mathbb{C}$ by the principle of analytic continuation. Moreover,

$$\eta(s) =\lim_{x\to-1^+}\sum_{n=1}^\infty\frac{x^{n-1}}{n^s}. \tag{1}$$

Indeed, Abel's theorem immediately establish this for $s > 0$, where $\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$ holds, and a more delicate argument shows that this identity continues to hold for all of $s\in\mathbb{C}$. Using this, we get

$$\zeta(0) =-\lim_{x\to-1^+}\sum_{n=1}^\infty x^{n-1} =-\lim_{x\to-1^+}\frac{1}{1-x} =-\frac{1}{2}$$

$$\zeta(-1) =-\frac{1}{3}\lim_{x\to-1^+}\sum_{n=1}^{\infty} nx^{n-1} =-\frac{1}{3}\lim_{x\to-1^+}\frac{1}{(1-x)^2} =-\frac{1}{12}$$

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$$\zeta(s) =\sum_{n=1}^\infty n^{-s}=\sum_{n=1}^\infty s\int_1^\infty x^{-s}dx=s \int_1^\infty \lfloor x\rfloor x^{-s-1}dx= \frac{s}{s-1}+\frac12-s \int_1^\infty (x-\lfloor x\rfloor-\frac12)x^{-s-1}dx$$

$$B_0(x)=x-\frac12,\qquad {B_{k+1}}'(x)=B_k(x),\qquad\int_1^2 B_{k+1}(x)dx=0$$

By induction starting with $k=0$, for $\Re(s) > -k$ $$\zeta(s)=\frac{s}{s-1}-\sum_{m=0}^k B_m(1)(\prod_{l=0}^m (s+l))- (\prod_{l=0}^k (s+l)) \int_1^\infty B_k(x-\lfloor x\rfloor)x^{-s-1-k}dx$$ (integration by parts) $$ =\frac{s}{s-1}-\sum_{m=0}^{k+1}B_m(1)(\prod_{l=0}^m (s+l))-( \prod_{l=0}^{k+1} (s+l) )\int_1^\infty B_{k+1}(x-\lfloor x\rfloor)x^{-s-1-(k+1)}dx$$

Whence

$$\zeta(-N)=\frac{-N}{-N-1}-\sum_{m=0}^N B_m(1)\prod_{l=0}^m (-N+l)$$

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  • $\begingroup$ Right. I don't guarantee there are no other mistakes, it is hard to play with such indices and I didn't check $\zeta(0),\zeta(-1)$, but the method involving Bernouilli is standard $\endgroup$ – reuns Sep 4 at 22:29
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A self-contained approach. One might start with $$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx $$ where the series and the integral are convergent over $\text{Re}(s)>1$. An analytic continuation over a larger subset of $\mathbb{C}$ can be achieved by applying IBP multiple times: for instance $$ \frac{1}{(s-1)\Gamma(s)}\int_{0}^{+\infty}x^{s-1}\cdot-\frac{d}{dx}\left(\frac{x}{e^x-1}\right)\,dx $$ provides an analytic continuation over $\text{Re}(s)>0$ and $$ \frac{(-1)^{k}}{(s-1)\Gamma(s+k+1)}\int_{0}^{+\infty}x^{s+k}\cdot\frac{d^{k+2}}{dx^{k+2}}\left(\frac{x}{e^x-1}\right)\,dx $$ provides an analytic continuation over $\text{Re}(s)>-(k+1)$. In particular, by picking $s=-k$ in the above line, $$ \zeta(-k)=\frac{(-1)^{k+1}}{k+1}\int_{0}^{+\infty}\frac{d^{k+2}}{dx^{k+2}}\left(\frac{x}{e^x-1}\right)\,dx =\frac{(-1)^k}{k+1}\cdot\left.\frac{d^{k+1}}{dx^{k+1}}\left(\frac{x}{e^x-1}\right)\right|_{x=0}$$ from which it follows that the values of the Riemann zeta function over the non-positive integers can be computed from the Maclaurin series of $\frac{x}{e^x-1}$. Since $$ \frac{x}{e^x-1}=-\frac{x}{2}+\underbrace{\frac{x}{2}\coth\frac{x}{2}}_{\text{even function}}$$ we have that $\zeta(-2n)=0$ for any $n\in\mathbb{N}^+$, and $\zeta(0)=-\frac{1}{2}$.

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  • $\begingroup$ @SimplyBeautifulArt: right, now fixed. $\endgroup$ – Jack D'Aurizio Sep 5 at 0:26
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    $\begingroup$ +1 very nice direct proof of the relation to the Bernoulli numbers via the coefficients of $x/(e^x-1)$'s Maclaurin series. $\endgroup$ – Simply Beautiful Art Sep 5 at 0:28

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