MarcL said:
I don't know it seems very broad to me. It's not the concept of a sum. It's when I was given the definition, that we use partial sum to verify if it goes to infinity. And I'm not even sure that is what my definition means. Fact is, and I think it is like that for mostly everybody, I hate applying methods / formula to concept I don't understand in general. I can solve those textbook problems but the concept of it is just... "fuzzy"
It's *not* just a sum. It's conceptually very sum-like, though.
As you have probably noticed, you cannot sum an infinitely many things. It takes too long (because it takes literally forever).
The best you can do is approximate.
You will notice that for *certain* series, they 'tend towards' some ultimate value. What you seem to be interested in is how to formalize the notion of 'tend towards'.
This is the concept of a limit, and it's the central notion in calculus. There are many different kinds of limits, but they all follow the same informal pattern:
"My boss asked me to approximate this with a certain degree of accuracy. How can I make sure I achieve that accuracy?"
In slightly less formal terms:
"If ε is the amount of error I'm allowed in my calculation, how can I guarantee that I always within that tolerance?"
In the case of a series, "how" you guarantee this is a question of how many terms in the series I have to add together. This how many is simply an integer (often called N).
To be slightly more precise, once we have added *at least* N terms, our error will be less than ε (regardless of how many terms total we add... the important thing, though, is we need it to work for at LEAST N terms).
So our final definition might look like this:
Let S be a series and let S_n be the partial sum of the first n terms of S. The we call a number L a limit of S iff for all positive real numbers ε, there exists an integer N, such that for all integers n such that n > N, we have S_n - L < ε.
For instance, suppose S is the series defined by S_n = 1/2 + 1/4 + 1/8 + ... + 1/(2^n).
We might play around with our calculator and guess the limit is 1. We then have to *prove it* (because conjectures need proofs!)
Proof lim S_n = 1.
Let ε be an arbitrary positive real. We need to find an integer N such that, for all integers n greater than N, 1 - S_n < ε.
With some clever arithmetic (or maybe simple induction), you can show that 1 - S_n = 1/(2^n). So our problem changes to finding that N such that for all n > N, 1/(2^n) < ε. Try rearranging that formula to 1 < ε * 2^n.
It should be clear if we have a fixed ε, we can always choose an n big enough to make this statement hold. And if we make n even bigger, it will continue to hold (meaning, it holds for all n > N). Thus, we have found our N.
Finding a satisfactory N was our goal, and so our proof is over. We proudly claim lim (1/2 + 1/4 + 1/8 + 1/16 + ...) = 1.
Some good facts about limits:
Limits do not always exist. The series 1 -1 + 1 - 1 + 1 - 1 + ... is an example of a series with no limit. We call these series "divergent".
If a series has a limit, then that limit is unique. (A big theme in math is existence and uniqueness. No only is there AT LEAST one solution... there is AT MOST one as well).
Doing finite sums is an algorithmic process -- anyone who knows how to add can get the right answer. However, doing infinite sums is an undecidable problem. You can't write a computer program (or hire a mathematician) that can always figure out (correctly) whether or not a series has a limit. In many simple cases, the problem is decidable (which is why Wolfram Alpha is so useful), but in general, if you think you have a process for solving limits, I could come up with a series that your program would not give the correct answer. (It would get stuck in an infinite loop or give me the wrong answer).
Cheers.