For a series to converge it must "tend to" or "approach" a number s. Mathematically there must be a natural number N for every real number \epsilon > 0 such that |s_n-s| < \epsilon for n \geq N, where s_n is the partial sum \sum^n_{k=1} a_k. In this case we say that the sum converges to s.
Note that it is important what you define as terms (the a_k's are the terms). 1-1+1-1+1-... is not the same as (1-1) + (1-1) + ...
The first sum has terms 1,-1,1,-1,... and so on, but the second sum has 1-1,1-1,1-1,... that is, 0,0,0,... as terms.
Obviously, if 1-1 = 0 are the terms, the series will converge to 0. The sum is simply 0 + 0 + 0 + ... which converges to 0 in the mathematical sense described above. Your example was 1 + (1-1) + (1-1) + ... which of course converges to 1. But this is not the same series.
On the other hand, if the terms are 1, -1, 1, -1, ... the series does not converge. And it is still not the same series as the other one. The reason is that it doesn't "tend to" or "approach" any number. The partial sum s_noscillates between 1 and 0. Mathematically, it doesn't satisfy the condition for convergence to any number s when we choose \epsilon = \frac{1}{2}. Can you see why?
Be careful around infinite series. Customary properties such as associativity and commutativity of terms doesn't apply in the same way.
