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.
Real Analysis Playlist: • Real Analysis
Intro to Sequences: • Intro to Sequences | C...
Definition of the Limit of a Sequence: • Definition of the Limi...
Sequences that Diverge to Infinity (Definition): • Sequences that Diverge...
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[@Kevbotxd]

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

[@WrathofMath]

Nice!

[@mahmoudalbahar1641]

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

[@WrathofMath]

Thank you, Mahmoud! I will keep going!

[@bhanugupta7750]

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

[@lalhriatpuiahmar5057]

But we say Epsilon greater than 0, so it can't be between - 1 & 0.

[@yashisingh6260]

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

[@wtt274]

Very clear explanation , thank you sir for this video !

[@WrathofMath]

Glad to help! Check out my analysis playlist for more and let me know if you have any questions! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@lalhriatpuiahmar5057]

Can we also use Epsilon = 1/3

[@rupsaseth2634]

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 .

[@maat6861]

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

[@turokg1578]

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?

[@nancyliu8145]

|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.

[@nishant0312]

Is it an oscillatory

[@kgopotsomasela4627]

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

[@majoorbaghdad1672]

2n-(-1)^n, prove diverges?

[@user-vn7ce5ig1z]

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

[@WrathofMath]

Nothing! Such is the nature of divergence. A divergent sequence is simply one that doesn't converge to any real number. Some of these happen to have the special property of "diverging to infinity" but others just oscillate!

[@user-vn7ce5ig1z]

@@WrathofMath Ah, okay, so a series that fluctuate back and forth between two or more values counts as "divergent"? I keep picturing the term as meaning it flys off _to_ somewhere. This makes sense now.

[@Davivlcp]

Perfect

[@WrathofMath]

Thank you! Check out my analysis playlist for more and let me know if you have any questions! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@jacksonmadison9994]

That wasn’t so bad