Bernoulli numbers, Riemann zeta, and strange sums

In the previous post, we looked at sums of the first n consecutive powers, i.e. sums of the form

where p was a positive integer. Here we look at what happens when we let p be a negative integer and we let n go to infinity. We'll learn more about Bernoulli numbers and we'll see what is meant by apparently absurd statements such as 1 + 2 + 3 + " = -1/12.

If p < -1, then the limit as n goes to infinity of S_p(n) is I(-p). That is, for s > 1, the Riemann zeta function I(s) is defined by

We don't have to limit ourselves to real numbers s > 1; the definition holds for complex numbers s with real part greater than 1. That'll be important below.

When s is a positive even number, there's a formula for I(s) in terms of the Bernoulli numbers:

The best-known special case of this formula is that

1 + 1/4 + 1/9 + 1/16 + " = I² / 6.

It's a famous open problem to find a closed-form expression for I(3) or any other odd argument.

The formula relating the zeta function and Bernoulli tells us a couple things about the Bernoulli numbers. First, for n a 1 the Bernoulli numbers with index 2n alternate sign. Second, by looking at the sum defining I(2n) we can see that it is approximately 1 for large n. This tells us that for large n, |B_2n| is approximately (2n)! / 2^2n-1 I²ⁿ.

We said above that the sum defining the Riemann zeta function is valid for complex numbers s with real part greater than 1. There is a unique analytic extension of the zeta function to the rest of the complex plane, except at s = 1. The zeta function is defined, for example, at negative integers, but the sum defining zeta in the half-plane Re(s) > 1 is NOT valid.

You may have seen the equation

1 + 2 + 3 + " = -1/12.

This is an abuse of notation. The sum on the left clearly diverges to infinity. But if the sum defining I(s) for Re(s) > 1 were valid for s = -1 (which it is not) then the left side would equal I(-1). The analytic continuation of I is valid at -1, and in fact I(-1) = -1/12. So the equation above is true if you interpret the left side, not as an ordinary sum, but as a way of writing I(-1). The same approach could be used to make sense of similar equations such as

1² + 2² + 3² + " = 0

and

1³ + 2³ + 3³ + " = 1/120.