# Sum ideas (part 4)

So how do you sum — or, if you prefer, associate a value with — $1+1+1+...$?

One way is zeta function regularization. For this we start with the sum

$\displaystyle S(s) = \sum_{k=1}^{\infty}k^{-s}$.

For example, $S(2) = 1+1/4+1/9+1/16+...$. This series converges to the limit $\pi^2/6$. In fact $S(s)$ converges for any complex $s$ where the real part $\Re(s) > 1$.

For $s=1$, $S(1) = 1+1/2+1/3+1/4...$ and this diverges. If you approach $s=1$ along the real axis you find $S(s)$ increases without limit. Off the real axis, things are a little different. At $s=1+i$, for example, the sum fails to converge, but as you approach t $s=1+i$ from the right, $S(s)$ approaches $0.58215 + 0.92684i$. Similar behavior is found elsewhere on the $\Re(s) = 1$ line, other than $s=1$.

That suggests there might be a way to construct a function that is equal to $S(s)$ for $\Re(s)>1$ but which has well defined values elsewhere, except $s=1$. And indeed there is: analytic continuation.

Imagine I give you the following function: $f(x) = x$ for real $x \in [0,1]$. Outside that interval $f(x)$ is undefined. But you obviously could define another function $g(x) = x$ which is defined on the whole real number line and has the property that $g(x) = f(x)$ in the range where $f(x)$ is defined. Obviously $g(x)$ is continuous, and is differentiable everywhere.

On the other hand, you could instead define $g(x)$ as being quadratics grafted onto the line from $(0,0)$ to $(1,1)$:

$g(x) = x^2+x \quad x < 0$
$= x \quad 0 \le x \le 1$
$= x^2-x+1 \quad x > 1$

which has the same properties. Or you could use cubics, or quartics, or, well, anything provided it has the right value and derivative at $0$ and $1$. There’s an infinite number of ways to continue $f(x)$ to the entire real line.

In the complex plane you can do something similar. I give you a function $f(z)$ defined for $z$ within some region of the complex plane. $f(z)$ is analytic, that is, it has a complex derivative everywhere it’s defined. Then you can give me an analytic function $g(z)$ defined everywhere in the complex plane and equal to $f(z)$ everywhere $f(z)$ is defined. (I’m being sloppy and informal here; there could be poles where neither function is defined, for example.)

Here’s the thing, though: Unlike $g(x)$ on the real line, $g(z)$ is unique. There is exactly one analytic function that continues my analytic function to the entire complex plane.

So, getting back to our sum (which is analytic), we can define an analytic function $\zeta(s) = S(s)$ for $\Re(s) > 1$, whose behavior for $\Re(s) \le 1$ is given by analytic continuation. One can show

$\displaystyle \zeta(s) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1} dx$

where $\Gamma(s) = \int_0^\infty x^{s-1}e^{-x} dx$ is the usual gamma function. $\zeta(s)$ has a pole at $s=1$ but is well defined everywhere else. $\zeta(s)$ is known as the Riemann zeta function.

Now, we know $\zeta(s)$ is the value of $\sum_{k=1}^{\infty}k^{-s}$ wherever that sum converges. Zeta regularization just assigns the value of $\zeta(s)$ to that sum where it does not converge as well. For instance, when $s=0$, we have $1+1+1+1+... = \zeta(0)$, and $\zeta(0) = -1/2$.

The somewhat notorious sum of the positive integers, $1+2+3+4+...$, is $S(-1)$, to which is assigned the value $\zeta(-1) = -1/12$. If you want to start an argument on the Internet, claiming that $1+2+3+4+... = -1/12$ is a good way to do it. Of course that claim glosses over a lot.

It turns out the negative even integers are (“trivial”) zeros of the zeta function, so $\sum_{k=1}^{\infty}k^{2n} = 0$ by this summation method. Generally, for integer exponents,

$\displaystyle \sum_{k=1}^{\infty}k^{n} = \zeta(-n) = (-1)^n \frac{B_{n+1}}{n+1}$,

where $B_n$ is the nth Bernoulli number,

$\displaystyle B_n = \sum_{k=0}^n\sum_{v=0}^k (-1)^v\binom{k}{v}\frac{v^m}{k+1}$.

So

$\displaystyle \sum_{k=1}^{\infty}k^{1} = -\frac{1}{12}$
$\displaystyle \sum_{k=1}^{\infty}k^{3} = \frac{1}{120}$
$\displaystyle \sum_{k=1}^{\infty}k^{5} = -\frac{1}{252}$
$\displaystyle \sum_{k=1}^{\infty}k^{7} = \frac{1}{240}$
$\displaystyle \sum_{k=1}^{\infty}k^{9} = -\frac{1}{132}$
$\displaystyle \sum_{k=1}^{\infty}k^{11} = \frac{691}{32760}$

and on from there.