Speaking of divergent series, I’d like to mention one of my favorites bits of informal folklore: this starts with a slogan
The only “truly” divergent series is the harmonic series
$latex H=1+\frac{1}{2}+\cdots +\frac{1}{n}+\cdots,$
which, of course, deserves plenty of quotes. But the idea is quite sound: it is that, given about any other series with real or complex terms (which does not involve this one in a hidden manner, of course, e.g., as an “irreducible summand”), such as
$latex \sum_{m\geq 1}{(-1)^{m+1}\frac{m^2}{4m^2-1}},$
it is quite often possible, by some kind of trick or feint, to assign a “value” (i.e., a real or complex number) to the series in a way which is reasonable and useful for certain purposes. [A famous example is the (apparently) ultra-divergent series
$latex \sum_{m\geq 0}{(-1)^m m!}=1-1+2-6+24-120+720-\cdots,$
which was considered by Wallis and Euler: its “value” is
$latex e\int_0^1{e^{-1/x}x^{-1}dx}=e\int_1^{+\infty}{y^{-1}e^{-y}dy}=0.5963473623231\ldots$
(see this article for an explanation of some of Euler’s ideas to do this “computation”).]
One setting in which the philosophy above has been refined to a precise (conjectural) statement is the theory of L-functions (over number fields). Indeed, observe that (formally) we have
$latex H=\zeta(1),$
where ζ(s) is the Riemann zeta function, which is only defined properly for the real part of s larger than one by the series
$latex \zeta(s)=\sum_{n\geq 1}{\frac{1}{n^s}}.$
Then, after translating the basic insight about the harmonic series from Dirichlet series to L-functions, one gets the following folklore conjecture:
If a reasonable (say, automorphic, or “motivic”) L-function L(s) over a number field k has a pole at s=1 (when normalized so that the functional equation relates values at s and 1-s), then L(s) is divisible by the Riemann zeta function, in the sense that
$latex L_1(s)=\frac{L(s)}{\zeta(s)}$
does not acquire poles at any of the complex zeros of the Riemann zeta function.
This is quite a rich problem: it contains a famous conjecture of Artin (the Dedekind zeta function of a number field should be divisible by the Riemann zeta function), and applied to the Rankin-Selberg convolution, it suggests the existence of the symmetric-square L-functions of modular forms — indeed, the first crucial result towards the “automorphic” existence of the latter (due to Gelbart and Jacquet), was the proof of this divisibility property by Shimura).
Now for the most amazing thing concerning this folklore property (at least to me): it seems that it also works modulo primes! Let me explain this (an explanation which I heard from Jean-Pierre Serre): another renowned formula of Euler for the values of the zeta function at even integers can be rephrased, after using the functional equation
$latex \pi^{-s/2}\Gamma(s/2)\zeta(s)=\pi^{-(1-s)/2}\Gamma((1-s)/2)\zeta(1-s),$
as the formula
$latex \zeta(1-2n)=1+2^{2n-1}+3^{2n-1}+\cdots + k^{2n-1}+\cdots =-\frac{1}{2n}B_{2n},$
where the Bernoulli numbers B2n are rational numbers defined by the power series expansion
$latex 1-\frac{1}{2}x+\sum_{n\geq 1}{B_{2n}\frac{x^n}{n!}}=\frac{x}{e^x-1}$
(and, of course, the middle expression for ζ(1-2n) is purely formal: it is one more example of a divergent series that can be given a convincing value).
The idea now is that the last expression (-1/2n B2n) can be reduced modulo a prime p, provided p does not divide 2n and does not divide the denominator of the Bernoulli number. Now, lo and behold, the primes dividing the denominators of Bernoulli numbers are known: they are exactly the primes such that
$latex p-1\mid 2n$
or equivalently such that
$latex k^{2n-1}\equiv k^{-1}\text{ mod } p$
for all integers k coprime with p (those for which the inverse modulo p makes sense…). So the Bernoulli number can not be reduced modulo primes exactly when (either p divides 2n or) formally we have
$latex -\frac{1}{2n}B_{2n}\equiv 1+2^{2n-1}+3^{2n-1}+\cdots + k^{2n-1}+\cdots\equiv 1+2^{-1}+\cdots +k^{-1}+\cdots,$
the divergent harmonic series again! (One must omit the terms with k divisible by p of course, but since this is purely formal, why not?).
I have no idea if there is a good explanation for this coincidence, but it is remarkably beautiful, and it certainly gives a convincing argument for the fact that the numerators of Bernoulli numbers are much more mysterious: so are the zeros of the Riemann zeta function, in comparison with its pole…)