There are different procedures for extracting a real number from a divergent series. We tend to say:
$$ 1 - 1 + 1 - 1 + 1 - \dots = \frac{1}{2} $$
and we use the Cesaro averages as a rule to . Lately I am personally of the school that the partial sums are of two values;
$$ \sum_{n = 0}^N (-1)^n = \left\{ \begin{array}{cl} 1 & n \text{ is odd} \\ 0 & n \text{ is even} \end{array} \right. $$
Then there is some type of limit, but the result is not a real number - an element of $\mathbb{R}$ - it is a distribution. We could try to say:
$$ \lim_{N \to \infty} \sum_{n = 0}^N (-1)^n \approx \frac{1}{2}\delta_{0} + \frac{1}{2}\delta_1$$
Divergent series come up as part of regluarization in physics, which are attempts to get finite numbers when the aren't any. As an avid reader of hep-th I can attest to physicsts at prominent institutions, casually assigning finite value to bad divergent sums and integrals.
The methods in Hardy's divergent series book or the theory of distributions will formalize such an intuition.
Many discussions end the conversation when they establish a particular series is divergent. Hardy's approach seem to study how the series diverges and establish a rate of divergence. Here is one of his result on the theta function. Let $x \notin \mathbb{Q}$, and $q = re^{2\pi i n x} \in \mathbb{D}$ then
$$ 1 + 2 \sum q^{n^2} = o\left( \sqrt[4]{\frac{1}{1-r}}\right) $$
Although I am slightly confused since if the continued fraction digits are bounded - $x = [a_0; a_1, a_2, \dots]$ with $|a_n| \leq M$ he shows:
$$ \left| 1 + 2 \sum q^{n^2} \right| \asymp
\sqrt[4]{\frac{1}{1-r}} $$
This is from a the collected papers of Hardy, [1]
This derivation is from page 4 of Hardy's divergent series:
\begin{eqnarray*} \sum (-1)^{n-1}\frac{1 - \cos n \theta}{n^2} &=& \sum \frac{(-1)^{n-1}}{n^2} \sum (-1)^k \frac{(n\theta)^{2k+2}}{(2k+2)!} \\
&=& \sum (-1)^k \frac{\theta^{2k+2}}{(2k+2)!} \sum \frac{(-1)^{n-1}}{n^{2k}} \\
&=& \frac{1}{2}\theta^2( 1 - 1 + 1 - 1 + \dots) \\
&=& \frac{\theta^2}{4} \end{eqnarray*}
Then Hardy begins to express concern this derivation can't hold in general:
$$ \sum (-1)^{n-1} \frac{f(\theta)}{n^2} = \frac{\pi^2}{2} a_0 + \frac{\theta^2}{2} a_1 $$
This is not true for $f(\theta) = a_0 + a_1 \theta^2 + a_2 \theta^4 + \dots$ so what gives?
Even though $\sum (-1)^n = \frac{1}{2}$ is clearly right in many circumstances, Hardy shows it can lead to false statements. How do we reject something which is so right and intuitive?
Lastly, Hardy shows the prime number theorem. The odds of an n-digit number being prime is $\frac{1}{\text{# of digits of } n}$. Or the average of the Mobius function is $0$... among square-free number half of them have an even number of factors and half of them have an odd number of factors:
$$ \frac{1}{N} \sum_{n \leq N} \mu(n) = o(1) $$
These are proven from tauberian theorems although there are several other steps involved.
