# Analytic continuation

I have received some skepticism that there are possibly other ways of assigning the sum of the natural numbers to a number other than -1/12 so I will try to be more precise. I thought it would be also useful to derive the analytic continuation of the zeta function, which I will do in a future post.  I will first give a simpler example to motivate the notion of analytic continuation. Consider the geometric series $1+s+s^2+s^3+\dots$. If $|s| < 1$ then we know that this series is equal to

$\frac{1}{1-s}$                (1)

Now, while the geometric series is only convergent and thus analytic inside the unit circle, (1) is defined everywhere in the complex plane except at $s=1$. So even though the sum doesn’t really exist outside of the domain of convergence, we can assign a number to it based on (1). For example, if we set $s=2$ we can make the assignment of $1 + 2 + 4 + 8 + \dots = -1$. So again, the sum of the powers of two doesn’t really equal -1, only (1) is defined at s=2. It’s just that the geometric series and (1) are the same function inside the domain of convergence. Now, it is true that the analytic continuation of a function is unique. However, although the value of -1 for $s=-1$ is the only value for the analytic continuation of the geometric series, that doesn’t mean that the sum of the powers of 2 needs to be uniquely assigned to negative one because the sum of the powers of 2 is not an analytic function. So if you could find some other series that is a function of some parameter $z$ that is analytic in some domain of convergence and happens to look like the sum of the powers of two for some $z$ value, and you can analytically continue the series to that value, then you would have another assignment.

Now consider my example from the previous post. Consider the series

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

This series is absolutely convergent for $s>1$.  Also note that if I set s=-1, I get

$\sum_{n=1}^\infty (n-1) = 0 +\sum_{n'=1}^\infty n' = 1 + 2 + 3 + \dots$

which is the sum of then natural numbers. Now, I can write (2) as

$\sum_{n=1}^\infty\left( \frac{1}{n^s}-\frac{1}{n^{s+1}}\right)$

and when the real part of s is greater than 1,  I can further write this as

$\sum_{n=1}^\infty\frac{1}{n^s}-\sum_{n=1}^\infty\frac{1}{n^{s+1}}=\zeta(s)-\zeta(s+1)$  (3)

All of these operations are perfectly fine as long as I’m in the domain of absolute convergence.  Now, as I will show in the next post, the analytic continuation of the zeta function to the negative integers is given by

$\zeta (-k) = -\frac{B_{k+1}}{k+1}$

where $B_k$ are the Bernoulli numbers, which is given by the Taylor expansion of

$\frac{x}{e^x-1} = \sum B_n \frac{x^n}{n!}$   (4)

The first few Bernoulli numbers are $B_0=1, B_1=-1/2, B_2 = 1/6$. Thus using this in (4) gives $\zeta(-1)=-1/12$. A similar proof will give $\zeta(0)=-1/2$.  Using this in (3) then gives the desired result that the sum of the natural numbers is (also) 5/12.

Now this is not to say that all assignments have the same physical value. I don’t know the details of how -1/12 is used in bosonic string theory but it is likely that the zeta function is crucial to the calculation.