# Divergent series

Les séries divergentes sont en général quelque chose de bien fatal et c’est une honte qu’on ose y fonder aucune démonstration. ("Divergent series are in general something fatal, and it is a disgrace to base any proof on them." Often translated as "Divergent series are an invention of the devil...")

N. H. Abel, letter to Holmboe, January 1826, reprinted in volume 2 of his collected papers.

#### Divergent series are used in physics

Even though, 1+2+3+... = –1/12, is not really true, the result can be arrived at through a number of different arguments. The simplest uses the term-by-term manipulations found at the bottom of Wikipedia: 1 +2 +3 +4 + .... This "result" has applications in physics: See Quantum mechanics/Casimir effect in one dimension, where a careful analysis shows that the "result" is actually achieved by taking the difference between two very large quantities, both related to this series. And, the physical analysis gives a plausible explanation of why neither term is actually infinity. Instead, both quantities are represented by very large numbers that are finite, due to the fact that at above some threshold energy, metal becomes transparent to X-rays. From a physics point of view, and for the Casimir effect, there is no mystery, except for the fact that this weird result (that Σ n = –1/12) seems to arise in a wide variety of different mathematical arguments.

Are these mathematical arguments valid or correct? It depends on how one defines "valid" or "correct". Higher mathematics is less about achieving the truth than it is about deducing the consequences of assumptions. Viewed in this way, the fact that such mathematics can be useful is astonishing. Also astonishing, is the fact that this calculation of the experimentally observable Casimir effect seems to suggest that empty space is full of energy in the form of light waves. Does this "prove" that space is not empty ... that these light waves actually exist? Certainly not. In fact, the goal of modern physics is not to prove that anything exists, but is instead restricted to merely making predictions that can be tested by experiments. What does or does not actually exist is more the subject of philosophy than of physics.

## Can two wrongs make a right?

This section attempts to illustrate how a divergent series can seem to solve a calculus problem, even though it contains two "errors" that mysteriously cancel each other.

### Statement of problem

Evaluate the function $f(x)$ at $x=2$ if the following information is known about the function and all its derivatives at $x=0$ :

$f(0)=1$ , and $f'(0)=1$ , and $f''(0)=2$ , and $f'''(0)=3\cdot 2$ , and $f''''(0)=4\cdot 3\cdot 2$ , and so forth...

Using $f^{(n)}$ to denote the n-th derivative, we require that the function obeys the following condition for all n:

$f^{(n)}(0)=n!$ ### Solution

Since 2 is a moderately small number and $f(0)=1$ , it is reasonable to seek a solution of the form,

$f(x)=1+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}...$ where the constants, ${a_{n}}$ , have yet to be determined. It is not hard to convince yourself that

$f(x)=1+x+x^{2}+x^{3}+...$ obeys the condition that the n-th derivative equals n! (at x=0). Setting $x=2$ yields:

$f(2)=1+2+4+8+16+...$ Note that $2f(2)=2+4+8+16+...$ which implies that $1+2f(2)=f(2)\,.$ Solving this equation for $f(2)$ yields the solution: $f(2)=-1$ ### Two flaws in the mathematics?

Apparently this solution contains two false steps:

• The desired function may be written in the form $f(x)=1+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+...$ • $1+2+4+8+16+...=-1$ ... or is it possible that both statements are somehow correct? In mathematics, a blatantly false statement can be changed into a true statement by redefining the meaning of "equals". See for example MATLAB/Divergent_series_investigations or this solution to the Casimir effect. The latter suggests that,

$1+2+3+4+...=\infty -{\frac {1}{12}}$ ### Verify solution

To verify that $f(2)=-1$ , note that

$f(x)={\frac {1}{1-x}}\Rightarrow f(2)=-1\,.$ This function satisfies the condition that at $x=0$ , the n-th derivative $f^{n}(x)$ must equal $n!$ .

$f(x=0)=(1-x)^{-1}=1$ $f'(x=0)=(-1)(1-x)^{-2}(-1)=(1-x)^{-2}=1$ $f''(x=0)=(-2)(1-x)^{-3}(-1)=2(1-x)^{-3}=2$ $f'''(x=0)=(-3)(2)(1-x)^{-4}(-1)=3\cdot 2(1-x)^{-4}=3\cdot 2$ ...

### The complex plane

The infinite series $f(x)=\sum x^{n}$ could have been converted to $f(x)=(1-x)^{-1}$ while $x$ was small enough for the series to converge, rendering it unnecessary to employ the divergent series. The two regions on the real axis, x<1 and x>1 seem less disconnected if the function is viewed on the complex plane because analytic continuation allows one to connect the regions x<1 and x>1 in a way that students familiar with real-variable functions might not realize. This is one of many reasons why calculus is far more interesting and beautiful if extended into the complex plane.