Kindly explain this thoroughly ?

Convergence and divergence of a sequence with the help of '#ϵ# ' definition of the notion of limits.

1 Answer
Dec 26, 2017

If there is an #L < oo# where, for all #epsilon > 0#, there exists a #delta# such that #n>=delta => abs(x_n-L) < epsilon,# then the sequence #x_n# converges to #L.# If such an #L# cannot be found, then the sequence diverges .

Explanation:

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.