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By definition of the Riemann Zeta Function, $$\zeta\left(\frac{1}{2}\right) = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}.$$ Since $\forall n \geq 1 : \frac{1}{\sqrt{n}} \geq \frac{1}{n}$, we have that for all $N \geq 1$, $$\sum_{n=1}^N\frac{1}{\sqrt{n}} \geq \sum_{n=1}^N \frac{1}{n},$$ but it is well known that $$\lim_{N\rightarrow\infty}\sum_{n=1}^N\frac{1}{n}=\infty,$$ so $\zeta\left(\frac{1}{2}\right)$ diverges by the comparison test.

In other words, $\zeta\left(\frac{1}{2}\right)$ should equal positive infinity, correct? If so, why do Maple, Mathematica, and Matlab all return a value of around $-1.4604$ when asked to numerically approximate this value? For example, see here.

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3  
The question becomes even more clear for $\zeta(-1) = \sum_{n=1}^\infty n = -\frac{1}{12}$. Why would a sum of positive numbers give a negative number? –  Fabian Mar 4 '11 at 20:00
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That is not the definition of the zeta function! –  quanta Apr 19 '11 at 11:36
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You cannot apply the formula $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ to $s=1/2$ because its hold for $\Re(s)>1$ only. You have to use its analytic continuation. –  Américo Tavares Apr 19 '11 at 16:35

3 Answers 3

up vote 18 down vote accepted

The analytic continuation of $\zeta(s)$ for $Re(s)>0$ is given by

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

So this is why WolframAlpha gives you

$$\zeta \left( \frac12 \right) = -(1+\sqrt{2}) \sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt{n}}.$$

See here for more details.

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To supplement Derek's answer, the Riemann zeta function $\zeta(s) = \sum_{n} n^{-s}$ was originally only considered for $s = \sigma + i t$, where $\sigma > 1$. Note that when $\sigma > 1$, the zeta function always converges absolutely. However, Riemann showed that the definition of the $\zeta$ function could be extended via analytic continuation to the whole complex plane (with a pole at $s = 1$). There, it satisfies the functional equation

$$ \Gamma(s/2) \pi^{-s/2} \zeta(s) = \Gamma\left(\frac{1-s}{2}\right) \pi^{- \frac{1-s}{2}}\zeta(1-s). $$

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We have such a thing even with geometric series (which has many things easier than the Dirichlet/zeta-series).
Consider $s_2=1+2+4+8+...+2^k+...$ and $s_3=3+9+27+81+...+3^k+...$ .
Although each term of $s_3$ is bigger that that of the same index k in $s_2$ (and both sums are diverging), the sum $s_2=-1 (={1 \over 1-2}) $ is bigger that $s_3 = -{3 \over 2} (={3 \over 1-3})$ (You may look for "geometric series/analytic continuation")

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Another way to put it: the values assigned to $s_2$ and $s_3$ are thought of as "antilimits" of the geometric series. –  J. M. Apr 19 '11 at 11:36

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