Why is $\sigma_1(0)$ not $-\frac{1}{12}$? The Eisenstein series $\mathbb{G}_2$ is given by
$$\mathbb{G}_2(z) = -\frac{1}{24} + \sum_{n=1}^\infty \sigma_1(n) q^n$$
with $q=e^{2\pi i z}$ and
$$\sigma_1(n):=\sum_{d\mid n} d$$
for $n\in\mathbb N$. That's why some authors define $\sigma_1(0):=-\frac{1}{24}$, since then $\mathbb{G}_2(z)$ reads as
$$\mathbb{G}_2(z) = \sum_{n=0}^\infty \sigma_1(n) q^n.$$
As you may already know the sum of all natural numbers is $-\frac{1}{12}$.
If we apply our definition of $\sigma_1(n)$ to $n=0$ we get
$$\sigma_1(0)=\sum_{d\mid 0} d = \sum_{d=1}^\infty d=-\frac{1}{12}.$$
So in this case the definition of $\sum_{d=1}^\infty d=-\frac{1}{12}$ is inappropriate by a factor of two (we'd rather have $-\frac{1}{24}$ here).
In math I'm used to the principle that everything goes well together. Here it doesn't. Do you have explanations for that? Can this issue be fixed somehow?
 A: The genuine issue is that the natural characterization/construction of the Eisenstein series, as $E_k(z)=\sum_\gamma 1/(cz+d)^k$ does not converge for $k=2$, so is created by analytic continuation ("Hecke summation") by holomorphically/meromorphically continuing $E_{s,k}(z)=\sum_\gamma {1\over (cz+d)^k\cdot |cz+d|^{2s}}$ to $s=0$. For some non-trivial reasons, evaluation of the Fourier coefficients commutes with meromorphic continuation... The point is that it's not an issue of "defining" the weight-two Eisenstein series in some way and then asking why the details are what they are. Rather, it is that the thing is determined by analytic continuation...
The very-secondary/artifactual impulses to try to "normalize" everything are not necessarily "bad", but, if not based on any structural or conceptual thing, basically have no weight at all, beyond "wishing for simplicity".
One underlying technical point is that the constant term of Eisenstein series $E_{s,k}$ has two terms, one of which vanishes for $k>2$ and $s=0$, ... but outside the range of convergence of the holomorphic Eis., all bets are off.
A: Here's the result I came up with. We re-define $\sigma_1(n)$ as:
$$\sigma_1(n) := \frac{1}{2}\sum_{d|n}d+\frac{n}{d}$$
Note that this definition preserves the original function since $\frac{n}{d}$ is another divisor and we compensate for adding each divisor twice by dividing the result by 2. Now, evaluating for 0 gives:
$$\sigma_1(0) = \frac{1}{2}\sum_{d|0} d+\frac{0}{d} = \frac{1}{2}\sum_{d=1}^{\infty}d = -\frac{1}{24}$$
For being in a situation where rigour is practically impossible, I think this works out alright and it does offer pleasant results. I hope you find it satisfactory.
