Sum of infinite divergent series I have learned that positive infinity plus negative infinity isn't equal to zero, it's an indeterminate form. However what happens if we subtract two infinite divergent series $\displaystyle{\sum_{n=1}^{\infty} n=\infty}$ ? Is it sill an indeterminate form or is it zero?
$$\sum_{n=1}^{\infty} n-\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty} (n-n)=\sum_{n=1}^{\infty} 0=0.$$

You should always be worried when working with infinite quantities, and by "worried", I mean that you should be extra careful.
In your particular case, the quantity $\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right)$ is an indeterminate, because it does not make sense to give it any particular value. The problem is that the value of your expression changes if we rearrange the terms. Notice that "$\sum_{n=1}^\infty n$" is a symbol, that stands for the limit of the partial sums. As such, when you write:
$$\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right)= \sum_{n=1}^\infty (n - n) = 0$$
you have implicitly made an arbitrary choice on how you are adding the corresponding partial sums. But we could make another choice so that:
$$\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right) = (1+2 - 1)+(3+4 - 2) + (5+6 -3) + \cdots $$
$$=\sum_{k=1}^\infty ((2k-1)+2k-k) = \sum_{k=1}^\infty (3k-1) = \infty$$
which clearly diverges to infinity. Or we could have chosen another way:
$$\left(\sum_{n=1}^\infty n\right) - \left(\sum_{n=1}^\infty n\right) = (1 - 1-2)+(2 - 3-4) + (3 -5-6) + \cdots $$
$$=\sum_{k=1}^\infty (k-(2k-1)-2k) = \sum_{k=1}^\infty -(3k-1) = -\infty.$$
Since we cannot decide which one of the three possibilities is right (all three are simply arbitrary rearrangements of partial sums), we cannot give this quantity a determined value, and hence we call it an indeterminate.
