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?
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.
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$$ \prod_{c=1}^{\infty}(1-c/n)=1-1/(-12n)+1/(2*(-12n)^2)-1/6*(1/(-12n)^3+2*1/(120*n^3))+...$$