# Meaning of equality in zeta regularization

It is known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = \infty$$ but it is also known that $$\sum\limits_{n = 1}^\infty{n = 1 + 2 + 3 + \cdots} = -\frac{1}{{12}}$$ which can obtained using the analytic continuation of $\zeta(s)$. My question is: What is the true meaning of equality here? It is a common practice to write $f = {\mathcal O}(g)$ to mean $f \in {\mathcal O}(g)$. Is this something similar?

• Check this out. mathoverflow.net/questions/94512/… – Alexander Gruber Dec 6 '12 at 1:28
• it means zeta regularization! – user51427 Dec 6 '12 at 1:44
• – Will Jagy Dec 6 '12 at 1:59
• I'd say the second equality is metaphorical. – anon Dec 6 '12 at 2:21
• @Alex It is not, because my question is about the meaning of $=$. I know why it is true, but the use of $=$ does not make sense. – glebovg Dec 6 '12 at 2:25

One thing that is sometimes lost when talking about series is what the symbol $$\tag{1}\sum_{n=1}^\infty a_n$$ means.

The point is that it is defined as a limit, i.e. the usual definition of the symbol $(1)$ is that it is the limit (if it exists) of the sequence of partial sums. This is already subtle, because the notion of limit depends on the topology. The topology is canonical on $\mathbb R$ or $\mathbb C$, but not in normed vector spaces where one can still consider series. This definition also stumbles with the case of conditionally convergent series, where any other limit can be achieved by reordering.

Anyway, the definition of $(1)$ as the limit of the sequence of partial sums (which is $\infty$ in your example) is not the only possible definition. There are other notions, such as Cesàro-sum and Abel-sum (these two the most noteworthy among others).

Yet another meaning one can given to the $\sum\limits_{n=1}^\infty n$ is the analytic continuation you mention. This is an example of the so called Ramanujan* summation, where the number $-1/12$ is obtained for the series in the question.

In Cesàro, you define $$\sum_{n=1}^\infty a_n=\lim_{N\to\infty}\frac{s_1+\cdots+s_N}N$$ (if it exists), where $s_k=\sum\limits_{n=1}^k a_n$. Summable in the canonical sense implies Cesàro summable, with the same limit, but not vice versa.

The other canonical summability is the Abel one: you define $$\sum_{n=1}^\infty a_n=\lim_{x\to1^-}\sum_{n=1}^\infty a_nx^n.$$ Again, Cesàro summability implies Abel summability to the same limit, but not vice versa.

To see an example, consider $a_n=(-1)^n$. The usual series doesn't exist. But, using Cesàro or Abel, one can say that $\sum\limits_{n=1}^\infty(-1)^n=\frac12$.

• Nice answer. So the equality is justified, but the equality depends on the definition? Can you add something about the Ramanujan summation? – glebovg Dec 6 '12 at 2:17
• I just added a mention of the Ramanujan summation with a link to the Wikipedia article which certainly explains it better than I could. – Martin Argerami Dec 6 '12 at 2:48
• Since $=$ is symmetric and antisymmetric, should we not adopt the use of a different symbol to avoid ambiguity? – glebovg Dec 6 '12 at 20:10
• It's not about the equal sign. It's about the meaning of what's on the left side. What should be done is use different symbols for the different series, say $\sum_c$, $\sum_a$, and $\sum_r$ for the Cesaro, Abel, and Ramanujan sums. But no one is doing, because usually context is enough. – Martin Argerami Dec 6 '12 at 21:55
• There's also Borel summability... – J. M. is a poor mathematician Apr 9 '13 at 2:00