Can divergent series be useful? As explained by Terence Tao on his blog for example, it is possible to give a value to some divergent series using analytic continuation. For instance, that allows identities like
$$\sum\limits_{n \geq 1} 1 =  - \frac{1}{2}, \ \ \sum\limits_{n \geq 1} n = - \frac{1}{12} \ \ \text{and} \ \ \sum\limits_{n \geq 1} n^2 = 0.$$
My question is: 

Is it just an artificial definition or can it be really used in some computation or argument?

 A: Stirling's approximation, so useful for approximating $n!$, is actually a divergent series.  The terms decrease initially, but for a given $n$ they start to increase after a while.  The error of the approximation is bounded by the first ignored term, so if you want the best for a given $n$ you just keep adding until they start to grow.  For most purposes the first term or two is all you need.
A: One thing to remember is that the infinite summation is not simply repeated addition: it crucially involves something extra to make sense.
The definition of infinite sum one uses in calculus is designed to facilitate limit-based arguments; but that is not always what one is thinking of with an infinite sum.
If we were working with formal power series, for example, we have facts such as
$$ \sum_{n=0}^{+\infty} n x^n = \frac{x}{(x-1)^2} $$
which have nothing to do with convergence: if you multiply the left hand side by $(x-1)^2$ and simplify, all of the terms of the series cancel except for the $x$. It would thus be reasonable as shorthand to say, for example,
$$ \sum_{n=0}^{+\infty} n 2^n = 2 $$
And to make general arguments like this rigorous, we can define a new summation operator 
$$ \sum_{n=0}^{+\infty}' a_n := f(1) $$
whenever $f(x)$ is a meromorphic function satisfying
$$ \sum_{n=0}^{+\infty} a_n x^n = f(x) $$
in a neighborhood of $x=0$.
In other settings, such as the one in your link, series of the sort
$$ \sum_{n=1}^{+\infty} a_n n^s $$
are important, giving rise to other summation operators related to evaluating such series at values of $s$.
A: In addation to the post above:
You might reconise this as stirlings appoximation but it is actualy the first term of a divergent product.
$$n!= (n/e)^n  \sqrt{2n\pi}  \prod_{c=1}^{\infty}(1-c/n)$$
$$\prod_{c=1}^{\infty}(1-c/n)=e^{\sum_{c=1}^{\infty} ln(1-c/n)}$$
$$ e^{\sum_{c=1}^{\infty}\frac{\zeta(-c)(-1)^c}{(cn^c)}}$$
$$n!= (n/e)^n  \sqrt{2n\pi}   e^{\sum_{c=1}^{\infty}\frac{\zeta(-c)(-1)^c}{(cn^c)}}$$
This is the most ugly way to compute it, but also the most widely "accepted" i've the feeling. For me to just write out the product as sums is the most natural way.
$$\prod_{c=1}^{\infty}(1-c/n)=$$
$$1+\sum_{c=1}^{\infty}-c/n +$$
$$1/2!((\sum_{c=1}^{\infty}-c/n)^2-\sum_{c=1}^{\infty}(-c/n)^2)+$$
$$1/3!((\sum_{c=1}^{\infty}-c/n)^3-3*(\sum_{c=1}^{\infty}-c/n)*\sum_{c=1}^{\infty}(-c/n)^2+2*\sum_{c=1}^{\infty}(-c/n)^3)+... $$ 
And you will get$$ \prod_{c=1}^{\infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$
