When we think about a sequence converging, we usually imagine the terms getting closer and closer to some finite value. For example, in the sequence 1," "1 1/2," "1 3/4," "1 7/8," "1 15/16, ..., the terms get closer and closer to 2. If we call the first term x_0=1, then general form for a term in this sequence is x_n=2-1/2^n, and as we look at terms further into the sequence (i.e. as n approaches infinity), the 1/2^n piece of the expression gets smaller and smaller (i.e. 1/2^n approaches 0) with every step, leaving the 2 as the dominant part.
In fact, it is possible to say that there's a value for n, say delta, where 2-1/2^n is "close enough" to 2 for all n >= delta that the difference isn't significant. But then we should define what we mean by "significant."
Let's say the difference isn't significant if it's less than epsilon = 0.001. Then the value of delta which makes all further terms in the sequence "close enough to 2" for our liking is delta=10, since 1/2^10~~0.000977, which is less than our chosen epsilon. If we choose a smaller epsilon like 0.00001, then delta=17, since 1/2^17~~0.00000763.
That's what we mean by the epsilon"-"delta notion of sequence limits. For any significance level epsilon of our choosing, if we can find a delta such that the expression is within epsilon of a finite "limit" L for all n>=delta, then the sequence is said to converge to L.
It's like a 2-player game:
- A limit L is chosen and fixed.
- Alice thinks of a value for epsilon.
- Bob tries to find a delta such that the delta^"th" term (and all terms after it) are less than epsilon away from L.
If Bob can always find a matching delta for Alice's epsilon, no matter how small that epsilon is, then Bob wins (the sequence converges to L). If Alice can ever find an epsilon for which Bob cannot find a delta, then Alice wins (the sequence does not converge to L).
Note how I didn't say "the sequence diverges." That's because it's still possible to play the game with the same sequence but a different L that may allow Bob to win. But, if there is no L < oo where, for all epsilon > 0, there exists a delta such that n>=delta => abs(x_n-L) < epsilon, then the sequence diverges.
For the above example, the sequence defined by x_n=2-1/2^n converges to 2, because no matter how small an epsilon we may choose, we can always find a delta such that the delta^"th" term (and all further terms) are within epsilon of 2.
