The sum clearly diverges. This is indisputable. The point of the claim above, that
[tex]1+2+3+\cdots=-\dfrac1{12}[/tex]
is to demonstrate that a sum of infinitely many terms can be manipulated in a variety of ways to end up with a contradictory result. It's an artifact of trying to do computations with an infinite number of terms.
The mathematician Srinivasa Ramanujan famously demonstrated the above as follows: Suppose the series converges to some constant, call it [tex]C[/tex]. Then
[tex]\begin{matrix}C&=&1&+2&+3&+4&+5&+6&+\cdots\\4C&=&&+4&&+8&&+12&+\cdots\\-3C&=&1&-2&+3&-4&+5&-6&+\cdots\end{matrix}[/tex]
Now, recall the geometric power series
[tex]\displaystyle\sum_{n\ge0}x^n=1+x+x^2+x^3+\cdots=\dfrac1{1-x}[/tex]
which holds for any [tex]|x|<1[/tex]. It has derivative
[tex]\displaystyle\sum_{n\ge1}nx^{n-1}=1+2x+3x^2+4x^3+\cdots=\dfrac1{(1-x)^2}[/tex]
Taking [tex]x=-1[/tex], we end up with
[tex]1+2(-1)+3(-1)^2+4(-1)^3+\cdots=1-2+3-4+\cdots=\dfrac14[/tex]
and so
[tex]-3C=\dfrac14\implies C=-\dfrac1{12}[/tex]
But as mentioned above, neither power series converges unless [tex]|x|<1[/tex]. What Ramanujan did was to consider the sum [tex]1-2+3-4+\cdots[/tex] as a limit of the power series evaluated at [tex]x=-1[/tex]:
[tex]\displaystyle-3C=\lim_{x\to-1^+}\sum_{n\ge1}nx^{n-1}=\lim_{x\to-1^+}\frac1{(1-x)^2}=\frac14[/tex]
then arrived at the conclusion that [tex]C=-\dfrac1{12}[/tex].
But again, let's emphasize that this result is patently wrong, and only serves to demonstrate that one can't manipulate a sum of infinitely many terms like one would a sum of a finite number of terms.
