[Superiority] If lim(n→∞)xn=(lim) ̅(n→∞)xn=A, then A is the only convergence point, and {xn} is bounded.【The sufficiency is proved, that is, when the upper and lower limits of the sequence exist and are equal, it is denoted as A, proving that the sequence converges to A】
If there is ε00, there are infinite multiple terms of {xn} outside U(A, ε0), it is denoted as 【only if there are infinite multiple terms of the sequence outside A's neighborhood, the sequence may not converge, and the inverse method is still used】
x_(n1 ), x_(n2 ),…, x_(nk ),…, then {x_(nk )}Binding, [The original sequence is bounded, so the sub-column is also bounded, and this sub-column is an infinite sequence ]
is known from "there is infinite bounded sequence has clustering points", {x_(nk)} has clustering points B≠A.
B is also a clustering point of {xn}. Contradiction!
∴∀ε0, in U(A, There are only finite multiple terms with {xn} outside ε). [This is the necessary and sufficient conditions for the convergence of the sequence]
∴lim(n→∞)xn=A.
The following figure may help you better understand the relationship between the upper limit, the lower limit and the converged sequence:
Many people's misunderstandings about the limit and converged points of sequence comes from, and confusion with the natural sequence represented by n, that is, the limit problem of the sequence is explored on the vertical axis, not on the horizontal axis. horizontal axis just gives a definition domain. As can be seen in the figure, the point on the left, no matter how discrete, will not change the essence of the sequence limit and the gathering point. The key is that in the interval tending to infinity, A0 and A1 are the two gathering points of the sequence. If there are only these two gathering points, then A1 is the upper limit and A0 is the lower limit. Or there are other gathering points between A1 and A0, and they are still the upper and lower limits. And when A1=A2, it is obvious that this convergence point is the limit of the function. Do you understand?
