Proof: Sequence (-1)^n Diverges | Real Analysis

We prove the sequence (-1)^n diverges. This is an example of a sequence that diverges, but not to positive or negative infinity. Thus, we will prove it diverges using the plain old definition of divergence - that is, that it does not converge to any real number. To prove (-1)^n doesn't converge to any real number, we will assume for contradiction that it does converge to a real number, and show this forces a contradiction.

In a nutshell, by assuming (-1)^n converges to a limit L, this forces the terms of the sequence to get arbitrarily close to L. Since the sequence alternates between -1 and 1, this means -1 should be arbitrarily close to L and so should 1. This is obviously not possible. For example, 1 and -1 should both be within 1/2 of L, but this means L is both positive and negative. We'll see the details of this contradiction in the full lesson.

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[Kevbotxd]

Watched this before my Real analysis midterm and this was one of the questions!!!! LETS

[mahmoudalbahar]

Thank you for this good are cool mathematician going making great videos.

[anonymousalias]

Are there any functions of x that use the basic 4 functions (addition, addition but negative, division, and division of divided numbers) that follows this pattern?

Specifically I'm doing hijinks with (-1)^n toward a pattern of 3 that involve recursion, and I don't know a way to simplify powers of this when formulas including (-1)^n are in the power. As such, I'm looking for alternatives since all I need is a pattern where odds are all one constant and evens are all a different constant.

[turokg]

if i take two subsequences a_2n and a_(2n-1) since the two subsequences converge to different values can i say that it does not converge?

[bhanugupta]

Great video. Really appreciate it. They are very helpful.
I think we can take any value for epsilon between -1 and 1 .

[Amira-xr]

Thank you so much! I have been struggling in my Analysis classes but ever since I've found your channel everything is so much easier.

[kgopotsomasela]

7:16 can we set epilson to be a distance 0 < ε < 1??

[maat]

Do you have a divergence proof example for trig functions, since they oscillate? For example sin(npi/2).

[nancyliu]

|X(n)-X(n-1)|=|X(n)-A+A-X(n-1)|, which is smaller or equal to |Xn-A|+|X(n-1)-A |→0, so 2 is smaller or equal to 0.

[rupsaseth]

We can take epsilon upto 0.9 or 9/10 but cannot take it as 1 cause for epsilon = 1, the intervals can have a common point 0, which becomes the limit. We can say the limit does not exist or the function diverges as long as we do not have a common point between the 2 intervals, which in turn is possible when : interval 1 :- -1- 0. 9= -1. 9 to -1+0.9 = -0. 1 and interval 2:- 1-0.9 = 0.1 to 1+0.9 = 1.9, where we have no common points between the intervals. It is basically possible if we take epsilon = 0.1, 0.2, 0.3, ...., 0.9 .

[wtt]

Very clear explanation, thank you sir for this video !

[yashisingh]

.Thank u, my all doubts get solved with this wonderful lecture.

[vnceigz]

It doesn't converge to any number but it also doesn't diverge to +/-infinity, so what the heck does it diverge to? 😕