Analytic Continuation

How do we evaluate the following sum?

$1 + 2 + 4 + 8 + \cdots$

One approach would be to use the $p$ -adic metric. However, this strategy requires us to redefine our number system, and work with the metric completion $\bar{\mathbb Q}_p$ instead of $\mathbb R$ . But can we stay in $\mathbb R$ and $\mathbb C$ ?

Consider the geometric series below:

$\displaystyle 1 + z + z^2 + z^3 + \cdots = \frac 1{1-z},\quad z \neq 1.$

We observe that the left-hand side converges for $|z| < 1$ and diverges for $|z| \geq 1$ . The right-hand side isn’t all-powerful; it has a pole at $z = 1$ . However, it is well-defined for any $z \neq 1$ , even at $z = -1$ . If we do not interpret the left-hand side as the limit of partial sums, would we still be justified to say that the following equation holds?

$\begin{aligned} 1 - 1 + 1 - 1 + \cdots &= 1 + (-1) + (-1)^2 + \cdots \\ &= \frac 1{1-(-1)} \\ &= 1/2 \end{aligned}$

Here’s the idea: if we were allowed to define

$\displaystyle 1 + z + z^2 + z^3 + \cdots = g(z), \quad z \neq 1,$

we know that $g(z) = 1/(1-z)$ for $|z| < 1$ . Would it be possible if $g(z) \neq 1/(1-z)$ for $|z| \geq 1$ ?

It turns out that if $g$ were complex-analytic, then this alternative would be impossible.

Lemma 1. Suppose $f : \bar{B}_R(w) \to \mathbb C$ is a holomorphic nonzero function. Then there exists $r \in (0, R)$ such that $f(B_r(w) \backslash \{w\}) \subseteq \mathbb C \backslash \{0\}$ .

Proof. There are two cases: either $f(w) \neq 0$ or $f(w) = 0$ . The former implies the conclusion immediately thanks to the continuity of $f$ . Suppose the latter. By complex-analyticity, we can expand $f$ as a Taylor series about $w$ :

$\displaystyle f(z) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} \cdot (z-w)^k.$

Since $f \neq 0$ , there exists some smallest $m$ such that $f^{(m)}(0) \neq 0$ :

$\displaystyle f(z) = (z-w)^m \cdot \underbrace{ \sum_{k=0}^\infty \frac{f^{(k+m)}(0)}{(k+m)!} \cdot (z-w)^{k} }_{g(z)}.$

Since $g \neq 0$ , by the first case, there exists $r \in (0, R)$ such that

$g(B_r(w) \backslash \{w\}) \subseteq \mathbb C\backslash \{0\}.$

Since $|z-w|^m > 0$ for $z \neq 0$ , we also have

$f(B_r(w) \backslash \{w\}) \subseteq \mathbb C\backslash \{0\}.$

Theorem 1 (Identity Theorem). Fix holomorphic functions $f , g : D \to \mathbb C$ . If the closed set

$K := (f-g)^{-1}(\{0\}) \subseteq D$

contains some $u \in \bar D$ , then $K = D$ .

Proof. Define $h := f - g$ for simplicity. Since $K \cap \bar D$ is closed, there exists a sequence $\{u_n \} \subseteq K \cap \bar D \backslash \{u\}$ such that $u_n \to u$ .

We claim that there exists $\epsilon > 0$ such that $h(\bar B_\epsilon(u)) = \{0\}$ . Otherwise, for any $\epsilon > 0$ , $h|_{\bar B_\epsilon(u)} \neq 0$ . By Lemma 1, there exists $r \in (0, R)$ such that $h(B_r(u) \backslash \{u\}) \subseteq \mathbb C \backslash \{0\}$ . Since $u_n \to u$ , there exists sufficiently large $n$ such that $u_n \in B_r(u) \backslash \{u\}$ . By construction, $0 = h(u_n) \neq 0$ , a contradiction.

Now fix $v \notin \bar B_\epsilon(u)$ . Since $D$ is path-connected, let $\gamma : [0, 1] \to D$ denote any path from $\gamma(0) = u$ to $\gamma(1) = v$ . Then $\tilde h := h \circ \gamma$ is continuous. Define the compact set

$T := \{ t \in [0, 1] : \tilde h |_{[0, t]} = 0\}.$

We claim that $T = [0, 1]$ by real induction, so that $h(v) = \tilde h (1) = 0$ .

By definition, $\tilde h (0) = 0$ , so that $0 \in T$ . Suppose $[0, x] \subseteq T$ .
Since $\gamma(x) \in \bar D$ , by the previous argument, there exists $\epsilon > 0$ such that $h(\bar B_\epsilon(\gamma(x))) = \{0\}$ . Therefore, there exists $\delta > 0$ such that $[x,x+\delta] \subseteq T$ .
Finally, suppose $[0, x) \subseteq T$ . Since $T$ is closed, $[0, x] = \overline{[0, x)} \subseteq \bar T = T$ .

Therefore, $T = [0, 1]$ , as required.

Corollary 1. For any holomorphic function $g : \mathbb C \backslash \{1\} \to \mathbb C$ such that

$\displaystyle 1 + z + z^2 + z^3 + \cdots = g(z), \quad |z| < 1,$

we must have $g(z) = 1/(1-z)$ . In particular, the following series are justified in the sense of analytic continuation:

$\begin{aligned} 1 + 2 + 4 + 8 + \cdots &= -1, \\ 1 - 1 + 1 - 1 + \cdots &= 1/2, \\ 1 - 2 + 3 - 4 + \cdots &= 1/4. \end{aligned}$

Proof. That $g(z) = 1/(1-z)$ follows from the identity theorem, so that

$\begin{aligned} 1 + 2 + 4 + 8 + \cdots = g(2) &= \frac 1{1-2} = -1, \\ 1 - 1 + 1 - 1 + \cdots = g(-1) &= \frac 1{1-(-1)} = 1/2. \end{aligned}$

For the final result, given $|z| < 1$ we differentiate on both sides:

$\displaystyle 1 + 2z + 3z^2 + 4z^3 + \cdots = \frac 1{(1-z)^2}.$

Now the left-hand side still converges for $|z| < 1$ by the ratio test:

$\displaystyle \lim_{n \to \infty} \left| \frac{(n+2) \cdot z^{n+1}}{(n+1) \cdot z^n} \right| = \lim_{n \to \infty} \left| \frac{n+2}{n+1} \right| \cdot |z| = 1 \cdot |z| = |z|.$

Therefore, the right-hand side is an analytic continuation of the left-hand side. Substituting $z = -1$ ,

$\displaystyle 1 - 2 + 3 - 4 + \cdots = \frac 1{(1-(-1))^2} = \frac 14.$

The challenge of analytic continuation still lies in demonstrating the existence of one such analytic function that agrees with our original sum.

It may seem bizarre that $1 + 2 + 3 + 4 + \cdots = -1/12$ , instead of the more intuitive result of $\infty$ . Here’s a somewhat clever argument. Define the sums

$\begin{aligned} S &:= 1 + 2 + 3 + 4 + \cdots, \\ T &:= 1 - 2 + 3 - 4 + \cdots. \end{aligned}$

Taking their difference,

$\begin{aligned} S-T &= 4 + 8 + 12 + 16 + \cdots \\ &= 4 \cdot (1+2+3+4+ \cdots) = 4S. \end{aligned}$

Applying Corollary 1,

$S = -\frac 13 \cdot T = -\frac 13 \cdot (1/4) = -1/12.$

But how do we know that our usual operations for adding convergent sums remain when adding divergent sums? We don’t!

But with analytic continuation, we recover this surprising result and more.

Example 1. In the sense of analytic continuation, we have the following series:

$\begin{aligned} 1 + 2 + 3 + 4 + \cdots &= - 1/12. \end{aligned}$

Proof. Using the Riemann zeta function, we can construct a holomorphic function $\zeta : \mathbb C \backslash \{1\} \to \mathbb C$ such that

$\displaystyle \sum_{n=1}^\infty \frac 1{n^s} = \zeta(s), \quad \mathrm{Re}(s) > 1$

and $\zeta(-1) = -1/12$ . For any other holomorphic function $g : \mathbb C \backslash \{1\} \to \mathbb C$ such that

$\displaystyle \sum_{n=1}^\infty \frac 1{n^s} = g(s), \quad \mathrm{Re}(s) > 1,$

the identity theorem tells us that $g = \zeta$ necessarily. Therefore, in the sense of analytic continuation,

$\displaystyle 1 + 2 + 3 + 4 + \cdots = \sum_{n=1}^\infty \frac 1{n^{-1}} = \zeta(-1) = -1/12.$

Remark 1. There are non-analytic continuation methods to recover the same series. Check out this video by Numberphile on how it works.

—Joel Kindiak, 7 Mar 26, 2021H

KindiakMath

Analytic Continuation

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Analytic Continuation

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