For nonpositive integers, the series does not converge, but via analytic continuation one can show that ζ(-n) = -B_{n+1}/(n+1) for n ≥ 0. In particular, ζ(-1) = -1/12.

A standard consequence of the analytic continuation of ζ(s) is the relation ζ(1-n) = -B_n/n for n > 1. Taking n = 2 gives ζ(-1) = -1/12.

From this formula, we can deduce special values of ζ(s) for nonpositive values of s: ζ(0) = 1/π lim x→1 cos(πx/2) ζ(x) = −1/2, ζ(−1) = −1/(2π^2) ζ(2) = −1/12, ζ(−2m) = 0, m a positive integer. This calculation appears in the context of extending the definition of the Riemann zeta function to the entire complex s‑plane and showing it is analytic everywhere except at s = 1, where it has a simple pole of residue 1.

The analytic continuation of the Riemann zeta function satisfies ζ(-n) = -B_{n+1}/(n+1) for n ≥ 0. Since B_2 = 1/6, this yields ζ(-1) = -1/12.

For nonpositive integers, the series does not converge, but via analytic continuation one can show that ζ(−n) = − B_{n+1}/(n+1) for n ≥ 0 (using the convention that B1 = 1/2). In particular, ζ vanishes at the negative even integers because B_m = 0 for all odd m other than 1. Another particular value is ζ(−1) = −1/12. This gives a pretext for assigning a finite value to the divergent series 1 + 2 + 3 + 4 + ⋯, which has been used in certain contexts (Ramanujan summation) such as string theory.

For s > 1, the series converges and ζ(s) > 1. Analytic continuation around the pole at s = 1 leads to a region of negative values, including ζ(−1) = −1/12. The benefit of introducing the Riemann zeta function is that it can be defined for other values of s by analytic continuation. One can then define the zeta-regularized sum of 1 + 2 + 3 + 4 + ⋯ to be ζ(−1). From this point, there are a few ways to prove that ζ(−1) = −1/12.

Thus we see that the values (4), (5), (6), (7) obtained by analytic continuation are nothing more than the constant terms of the asymptotic expansion of the smoothed partial sums. In fact the formula arises from the analytic continuation of the Riemann zeta function itself, and it tells us that (−1/12) is the value of the function ζ(s) at s = −1.

We prove that a series derived using Euler's transformation provides the analytic continuation of ζ(s) for all complex s ≠ 1. At negative integers the series becomes a finite sum whose value is given by an explicit formula for Bernoulli numbers. … Thus ζ(0) = −1/2 and, using Wallis's product for π/2, one obtains the values at negative integers consistent with the Bernoulli number formula ζ(−n) = −B_{n+1}/(n+1), which for n = 1 gives ζ(−1) = −1/12.

The Riemann zeta function for negative even integers is 0 (those are the trivial zeros of the Riemann zeta function). The Riemann zeta function for nonnegative even integers is given by ... Analytic continuation to the left of the critical strip: for σ < 0 we have ζ(s) = 2^s π^{s−1} sin(πs/2) Γ(1−s) ζ(1−s), Re(s) < 0. This functional equation defines the analytic continuation of ζ(s) to the left half-plane and allows evaluation at negative integers such as s = −1.

As defined above, the zeta function ζ(s) with a complex number s is defined for Re(s) > 1. However, ζ(s) has a unique analytic continuation to the entire complex plane, excluding the point s = 1, which corresponds to a simple pole with complex residue 1. For nonpositive integers, this analytic continuation yields the values ζ(−n) = −B_{n+1}/(n+1), giving in particular ζ(−1) = −1/12.

It follows that ζ extends to a meromorphic function on C, regular except for a simple pole at s = 1, and that this analytic continuation has simple zeros at the negative even integers −2, −4, −6, ... . In particular, ζ(-1) = -1/12.

Corollary 2. The Riemann zeta function ζ(s) has its analytic continuation defined on ℂ \ {1}, with a simple pole at s = 1 with residue 1. … Now we are safe to compute ζ(−1). But I believe, after these long computations of the analytical continuation, we can be confident enough to say that, when Re(s) ≤ 1, the Riemann zeta function ζ(s) absolutely cannot be immediately explained by its ordinary definition ∑_{n=1}^∞ n^{−s}; instead its values, such as ζ(−1) = −1/12, come from the analytically continued formula.

Answer Gravy: Figuring out that ζ(−1) = −1/12 takes a bit of work. You have to find an analytic continuation that covers s = −1, and then actually evaluate it. ζ(s) is defined as ∑ 1/n^s when s>1 and as the analytic continuation of that sum otherwise. The analytic continuation of a function is unique, so nailing down ζ(s) for s>1 is all you need to continue it out into the complex plane.

One amazing thing about functions of complex numbers is that if you know the function sufficiently well for some set of inputs, then (up to some technical details) you can know the value of the function everywhere else on the complex plane. This method of extending the definition of a function is known as analytic continuation. OK. So now we have a function ζ(s) that agrees with Euler's zeta function S(x) when you plug in values x>1. When you plug in values x≤1, the zeta function gives you a finite output. What value do you get when you plug x=−1 into the zeta function? You've guessed it: ζ(−1)=−1/12.

The Riemann zeta function is one of the most important objects in analytic number theory. … Specifically, one thing we know about the extended zeta function is ζ(−1) = −1/12. This is a statement about the continuation of ζ(s), not a direct fact about the sum ∑ 1/n^s. Remember, the definition of the zeta function on the left half of the plane is not defined directly from this sum; instead it comes from analytically continuing this sum beyond the domain where it converges.

In fact, the Riemann Zeta function can be analytically continued to the whole complex plane. A possible way ... This is the functional equation, proved for complex values of s in the strip −2 < Re(s) < −1, and it is valid in the whole complex plane by analytic continuation. ... The latest formula rewrites as ζ(s) = −2^s (s+1)(s+2) ∑ ... which provides values of ζ(s) beyond the half-plane Re(s) > 1, via analytic continuation.

Defining the Riemann zeta function ζ(s) for Re s > 1, are holomorphic in the whole s-plane and the series converges in that half-plane, and the values of this function coincide with the values of the zeta function in the half-plane Re s > 1. This result means that, via the equation (3), the zeta function has been analytically continued to the domain D, as far as to the imaginary axis. ... Charles Hermite has shown that the zeta function may be analytically continued to the whole s-plane except for a simple pole at s = 1, by using the equation ζ(s) = 1/Γ(s) ∫_0^∞ x^{s−1}/(e^x − 1) dx.

For negative integers, one has ζ(-n) = -B_{n+1}/(n+1). Thus ζ(-1) = -B_2/2 = -1/12, using B_2 = 1/6.

Our purpose in this chapter is to extend this definition to the entire complex s‑plane, and show that the Riemann zeta function is analytic everywhere except at s = 1, where it has a simple pole of residue 1. … From this formula, we can deduce special values of ζ(s) for nonpositive values of s, including ζ(−1) = −1/(2π^2) ζ(2) = −1/12. Thus, the value ζ(−1) = −1/12 arises naturally from the analytic continuation and the functional equation relating values of ζ at s and 1 − s.

The Riemann zeta-function ζ(s) is defined by ζ(s) := ∑_{n=1}^∞ 1/n^s (1) for Re s > 1, but it is well known that there exists an analytic continuation onto the whole complex plane except for a simple pole at s = 1. ... Theorem. ζ(s), as defined by (1), extends analytically onto the whole s-plane, except a simple pole at s = 1. Let Λ(s) := π^{−s/2} Γ(s/2) ζ(s). Then Λ(s) = Λ(1 − s). ... Thus we have found a meromorphic function on C with a (simple) pole at s = 1, and which equals ζ(s) for Re s > 1. This is our analytic continuation.

The Riemann zeta function evaluated on the real line at −1 yields the sum of natural numbers, but of course that is not too meaningful since the Riemann zeta function is only defined for Re(z) > 1. It turns out the analytic continuation of the Riemann zeta function evaluated on the real line at −1 is −1/12. Short answer: It's not −1/12 [as a direct sum], it diverges to infinity (like one would suspect), but via analytic continuation the value assigned to ζ(−1) is −1/12.

ζ(−1) = −1/12, but the sum of all natural numbers is absolutely not equal to −1/12. The sum of all natural numbers is a divergent series, and therefore doesn't equal any real (or complex) number. The series definition of the Riemann zeta function only converges for values z such that Re(z) > 1. To evaluate it at any other value, you have to use analytic continuation. The famous value of −1/12 comes up with the analytic continuation of Riemann's ζ function.

To prove that the Riemann Zeta Function has an analytic continuation, we find an analytic function on the entire complex plane (except s=1) that equals the zeta function when Re(s)>1. By the property I stated about analytic functions, this is the only such function, so it is the unique extension of the zeta function to the complex plane. ... This completed zeta function, Xi(s), has an analytic continuation and satisfies a simple reflection. Using these relations one can define ζ(s) for values such as negative integers where the original series does not converge.

In this lecture on analytic continuation and the zeta function, the speaker explains that the function ζ(s) defined by the series ∑_{n=1}^∞ n^{−s} does not even make sense for s < 1 because the series diverges there. … If we just pretend that this series definition of the zeta function makes sense at −1 then we would see that ζ(−1) should be 1 + 2 + 3 + 4 + … On the other hand, using analytic continuation we can just compute directly that ζ(−1) happens to be −1/12.

{ts:964} This process of extending an analytic function in the only way possible that's still analytic is called, as you may have guessed, analytic continuation. So that's how the full Riemann zeta function is defined. {ts:1166} Another thing we know about this extended function is that it maps the point negative one over to negative one twelfth. Instead, it comes from analytically continuing the sum beyond the domain where it converges.

In this video, I develop a more compact functional equation for the Riemann Zeta function and discuss its potential applications in number theory and physics. ... [Transcript] "the pi squares cancel out we have negative 1 over 12 and just look at the left hand side here it's 1 plus 2 plus 3 plus 4 plus 5 going on till infinity the sum of all positive integers or all natural numbers negative 1 over 12". The speaker is using the value ζ(−1) = −1/12 obtained via analytic continuation to symbolically assign −1/12 to the divergent series 1 + 2 + 3 + ⋯.

The function ζ(s) has important applications in mathematics and physics. In mathematics, ζ(s) is closely related to number theory, particularly the distribution of prime numbers. In physics, ζ(s) appears in quantum field theory and statistical mechanics. ... We review the standard constructions of the analytic continuation of the Riemann zeta function, emphasizing contour integral representations and functional equations that allow evaluation at negative integers. Although the power series definition diverges for Re(s)≤1, the analytically continued ζ(s) takes finite values such as ζ(0)=−1/2 and ζ(−1)=−1/12.