Today, as I was flipping through my copy of Higher Algebra by Barnard and Child, I came across a theorem which said,

The series $$ 1+\frac{1}{2^p} +\frac{1}{3^p}+...$$ diverges for $p\leq 1$ and converges for $p>1$.

But later I found out that the zeta function is defined for all complex values other than 1. Now I know that Riemann analytically continued this function to fit all complex values, but how do I explain, to a layman, that $\zeta(0)=1+1+1+...=-\frac{1}{2}$?

The Wiki articles on these topics go way over my head. I'd appreciate it if someone can explain it to me what analytic continuation actually is, and which functions can be analytically continued?


If the function diverges for $p\leq1$, how is WolframAlpha able to compute $\zeta(1/5)$? Shouldn't it give out infinity as the answer?

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    $\begingroup$ Do you know any complex analysis? If you don't, you should learn some complex analysis first. $\endgroup$ May 29, 2011 at 11:44
  • $\begingroup$ Well I don't know any complex analysis. The reason I ask this question is that I've stumbled upon a result and I want to see if I can "analytically continue" the fractional part function. (I really hope I'm not sounding terribly stupid.) $\endgroup$
    – kodyv
    May 29, 2011 at 11:48
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    $\begingroup$ @Koundinya: Everything can't be explained to a layman. IF everything can be explained to a layman, then all layman would be Mathematicians :) $\endgroup$
    – user9413
    May 29, 2011 at 12:01
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    $\begingroup$ Tell your layman these: $\zeta(0)=1+1+1+...$ is wrong. $1+1+1+...=-\frac{1}{2}$ is wrong. $\zeta(0)=-\frac{1}{2}$ is correct. $\endgroup$
    – GEdgar
    May 29, 2011 at 13:04
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    $\begingroup$ For $z\in\mathbb{C},\text{Re }z>1$ the zeta function can be expressed as $$\zeta (z)=\sum_{k=1}^{\infty }\frac{1}{k^{z}}=\frac{1}{1-2^{1-z}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{z}}.$$ Since the alternating series on the RHS converges for $\text{Re }z>0$, the function $\zeta (z)$ can be analitically extented to $\text{Re }z>0$ as $$\zeta (z)=\frac{1}{1-2^{1-z}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{z}}.$$ For $z=1/2$, we get $$\zeta (1/2)=-\left( 1+\sqrt{2}\right) \sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}.$$ $\endgroup$ May 29, 2011 at 13:27

3 Answers 3


I'll give you the world's simplest example. $1+x+x^2+\dots$ converges for $|x|\lt1$ only. The function $1/(1-x)$ is analytic everywhere except for a pole at $x=1$, and agrees with $1+x+x^2+\dots$ everywhere the latter is defined, so $1/(1-x)$ is the analytic continuation of $1+x+x^2+\dots$. In that sense, $1+2+4+\dots=-1$.

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    $\begingroup$ For a suitable definition of "+", anyway. $\endgroup$ May 29, 2011 at 13:40
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    $\begingroup$ Is there a unique analytic function for all or most sequences? $\endgroup$
    – Jus12
    Jul 25, 2015 at 15:15
  • $\begingroup$ I don't know what you mean by "sequences". There are no sequences in my post. $\endgroup$ Jul 25, 2015 at 23:06
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    $\begingroup$ @Jus, perhaps you are asking whether, given a sequence $a_0,a_1,\dots$, there is an analytic function $a_0+a_1z+a_2z^2+\cdots$ which has an analytic continuation to $z=1$. There are at least two obstacles. One, if the sequence grows too fast, the series won't converge at all for $z\ne0$, so you won't have an analytic function to continue. Two, even if the series does converge to an analytic function in some region, that region may have a "natural boundary" beyond which analytic continuation is impossible. $\endgroup$ Jul 26, 2015 at 0:16
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    $\begingroup$ @Alexander, $1+z^2+z^4+z^6+\cdots$ can be extended to the entire complex plane, excepting only $z=\pm1$; it's just $1/(1-z^2)$. Maybe you are thinking of $z+z^2+z^4+z^8+\cdots$, which doesn't converge anywhere on $|z|=1$, and can't be extended analytically beyond it. $\endgroup$ Apr 12, 2021 at 4:47

If all you're interested in is an explanation to a layman of why a thing like analytic continuation makes any sense to begin with (in particular, why it spits out a unique answer), the answer is the identity theorem for holomorphic functions, which in its stronger form says that if two holomorphic functions $f, g$ defined on a connected open subset $U$ of $\mathbb{C}$ are equal on a set of points in $U$ with an accumulation point (in particular any open subset of $U$), then in fact $f = g$. This is an extremely strong rigidity theorem generalizing the corresponding fact for polynomials, and shows that if you want to extend a holomorphic function $f$ defined on some domain $U$ to a function $\tilde{f}$ defined on some larger connected domain $V$, then there is at most one way to do it (since any two extensions agree on $U$ and hence agree on $V$).

One sense of "analytic continuation" is that it refers to any function $\tilde{f}$ with the above property, and another sense of "analytic continuation" is that it refers to methods for defining $\tilde{f}$ given $f$.

Sometimes it is the case that a function $f$ can be analytically continued to a very large domain $U$ but that a series or integral defining $f$ can't be made to converge on all of $U$. The philosophical point to take away from this is that holomorphic functions have a sort of "Platonic reality" that isn't necessarily perfectly captured by any particular series or integral definition of them, which are just imperfect "shadows" of this Platonic reality. The slogan I use in this situation and others like it is that

convergence is overrated.

In response to the edit, WolframAlpha is using "shadows" other than the standard series definition of the Riemann zeta function.

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    $\begingroup$ ... and apparently under-appreciated by some :) No need to explain that as an analytically inclined person, I don't like this slogan (but as any slogan only captures part of the truth, I can live with it)... $\endgroup$
    – t.b.
    May 29, 2011 at 12:34
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    $\begingroup$ @Theo: convergence is overrated when doing combinatorics. On reflection, it's useful there, but more often there you don't want to care bout convergence. $\endgroup$
    – Mitch
    May 29, 2011 at 15:23
  • $\begingroup$ What do you mean by Wolfram uses "shadows"? What is a "shadow"? $\endgroup$ Jun 27, 2019 at 6:42

Answer to the added paragraph only. I guess at least this part is simple. In the case when the argument of the zeta function is a real $x>1$ the zeta function can be expressed as $$\zeta (x)=\sum_{k=1}^{\infty }\frac{1}{k^{x}}=\frac{1}{1-2^{1-x}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}},\qquad(\ast)$$

where the last series is the Dirichlet eta function. Since for $x>0$ this alternating series converges (because $1/k^x\to 0$), and $1-2^{1-x}\neq 0$ for $x\in]0,1[\cup]1,\infty[$, the function $\zeta (x)$ can be continued analytically to $x\in]0,1[\cup]1,\infty[$ (note that $1$ is excluded as in the first series) as

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

For $x=1/2$, we get $\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}\approx 0.6049$ and

$$\zeta (1/2)=-\left( 1+\sqrt{2}\right) \sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}\approx -1.4604.$$

Added: Derivation of $(\ast)$:

$$\begin{eqnarray*} \zeta (x) &=&\sum_{k=1}^{\infty }\frac{1}{k^{x}}=\sum_{k=1}^{\infty }\frac{% (-1)^{k-1}}{k^{x}}+2\sum_{k=1}^{\infty }\frac{1}{\left( 2k\right) ^{x}} \\ &=&\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}}+2^{1-x}\zeta (x). \end{eqnarray*}$$

The relation is as follows.

  • $\begingroup$ I prefer to directly subtract the eta function out of the zeta function. Writing it out term by term this way is more visual for me. $\endgroup$ Dec 21, 2016 at 1:00

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