I can't understand that statement i heard during a math course with the fact that the partial sum's series of the natural numbers don't converge and the fact that according to Euler and Riemann the infinite sum of the naturals equals -1/12....
By the way, how come the latter result doesn't violate numbers theory: I suppose that if you sum natural numbers you can't obtain an element essentialy different from them as i believe a rational number is...I mean they're outside the naturals group! does it matter at all?
In terms of infinite series, an infinite sum is defined as the limit of the partial sums, and so, as you stated, the infinite sum converges if and only if the sequence of partial sums does, as they are the same thing. So, in this context, $\sum_{n=1}^{\infty} n$ diverges.
To say that $\sum_{n=1}^{\infty} n = \frac{-1}{12}$ is a little misleading, to say the least. This is often quoted in this form for its shock value, because we mathematicians like to keep an air of mystery about our work (mostly so that laypeople think we are smart ;-) ).
Note that we can write $\sum_{n=1}^{\infty} n$ as $\sum_{n=1}^{\infty} \frac{1}{n^{-1}}$. This may look like a strange thing to do, but in your study of series you will no doubt come across the Harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$, and the related series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ for any $p > 0$.
The Riemann zeta function, $\zeta(s)$, is an extension of this last generalisation, where the value $p$ is replaced by a complex number (and changes its name to $s$). Now the series $\sum_{n=1}^{\infty} \frac{1}{n^s}$ converges whenever $s$ is a real number greater than $1$. The zeta function, $\zeta(s)$, is what is called an analytic continuation of this function to the complex plane excluding $s = 1$ - it is a well-defined function taking values for all $s \in \mathbb{C} \setminus \{ 1 \}$ such that $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$ whenever $s$ is a real number bigger than $1$.
It just so happens that $\zeta(-1) = \frac{-1}{12}$, which leads to the oft-quoted statement that is not technically correct. It is incorrect because $\zeta(s)$ is not defined as $\sum_{n=1}^{\infty} \frac{1}{n^s}$ for all values of $s$.
Finally, one should note that this zeta function is very important in mathematics; the famous Riemann Hypothesis concerns (non-trivial) zeroes of this function, and is deeply tied to the distribution of prime numbers, among other things.
EDIT: As mentioned by @joriki, there are other approaches that lead to the value of $\frac{-1}{12}$, and are covered in this Wikipedia article.
ADDENDUM: It seems some additional remarks on analytic continuations might be helpful. While the Riemann zeta function itself is too complex (pun intended) for me to explain well here, I will take a simpler, and yet hopefully still illustrative, example.
Given non-negative integers $n,k \ge 0$, the binomial coefficient $\binom{n}{k}$ can (dare I say should) be defined as the number of subsets of size $k$ of an $n$-element set. Some elementary counting gives the formula $\binom{n}{k} = \frac{n!}{k! (n-k)!} = \frac{1}{k!} \prod_{j=0}^{k-1} (n - j)$.
Now when you look at this formula, there is no reason why $n$ should be an integer. Indeed, we can define a function $f(x) = \frac{1}{k!} \prod_{j=0}^{k-1} (x - j)$, which is defined for all $x \in \mathbb{R}$ (or even $x \in \mathbb{C}$, should you so wish). Indeed, this is a polynomial, which is about as nice as functions can get, and it has the property that whenever $x \in \mathbb{Z}_{\ge 0}$, $f(x) = \binom{x}{k}$ counts the number of subsets of size $k$ of an $x$-element set.
We can plug other values into this function too. For example, $$ f(-1) = \frac{1}{k!} \prod_{j=0}^{k-1} (-1 - j) = (-1)^k. $$ However, we cannot say that a set of size $-1$ has $(-1)^k$ subsets of size $k$; that is patently nonsense. The equivalence between that definition of the binomial coefficient and this polynomial function is only valid for $x \in \mathbb{Z}_{\ge 0}$.
Similarly, the zeta function $\zeta(s)$ is equal the series $\sum_{n = 1}^{\infty} \frac{1}{n^s}$ when $s \in \mathbb{R}_{> 1}$. However, it extends that series to a wider domain (namely $\mathbb{C} \setminus \{1\}$) to give a function with very nice analytic properties - this is what is meant by an analytic continuation. It so happens that $\zeta(-1) = \frac{-1}{12}$, but it is incorrect to claim that $$ \sum_{n=1}^{\infty} n = \sum_{n=1}^{\infty} \frac{1}{n^{-1}} = \zeta(-1) = \frac{-1}{12}, $$ since the second equality above is not valid for $s = -1$. As is easy to show, the series $\sum_{n=1}^{\infty} n$ diverges instead.