Example of Cauchy Sequence 1

Edit: Last term in summation at 4:30 should be 1/2^{n-1}, not 1/2^n. In the next line (top of board), the last term should be 1/2^{n-m}.

Real Analysis: Let {x_n} be a sequence of real number such that |x_n - x_{n+1}| lt 1/2^n for all n gt 0. Show that {x_n} is a Cauchy sequence. We focus on basic techniques for epsilon proofs.

[xoppa]

@MathDoctorBob  these lectures are much more clear and easy to see than the lectures by Professor Francis Su at Harvey Mudd College  (which are on youtube).
Thankyou so much. I would like to donate money if thats possible (and if you could make a whole curriculum that would be awesome).

[MathDoctorBob]

Thanks for the high praise! I keep promising to work on a full Real Analysis playlist. It is at the top of the queue - algebra keeps pulling me back in.

[DoneBefour]

I'm in Foundations of analysis at UTSA you just made this shit a whole lot easier thanks !

[integralmath]

I like the new pointer. I half expected you to start off with 'wingardium leviosa'.

[MathDoctorBob]

Thanks for the catch and the kind words! Triangle inequality allows equals; I'll annotate.

Nothing on Bolzano-Weierstrass beyond what is in calculus (monotonic + bounded = convergent). I may get to real analysis this summer. It depends on my teaching load.

[MathDoctorBob]

@RespectMyHate Math goes on forever. It truly is a universe unto itself.

You could probably take this right after a few calculus classes, but it's probably better to have a course in linear algebra or differential equations as a bridge to proofs. In those classes, you need to read proofs, but the writing is not intense. In real analysis, a good amount of time is spent specifically learning to write proofs.

Real analysis is next for me here. First we do proofs (BM sequence). - Bob

[MathDoctorBob]

+MrAlexSM09 (Wiped comments too soon) Here's the most efficient way I see for when |x_(n+1)-x_n|<1/n^2.

1) |x_(n+1) - x_(m)|<1/n^2 +...+1/m^2 by the triangle inequality

2) 1/m^2 < 1/(m-1)m = 1/(m-1)-1/m

3) Thus |x_(n+1) - x_m|<1/(m-1)-1/n (telescoping)<1/N+1/N<2e.

[MathDoctorBob]

Thanks! As noted below, Analysis this summer. Maybe.

[jameskluz]

Thanks Doc! These videos are always so helpful!

[jreylbc]

Why do you change your inequality from 1/2^m <= 1/2^n to 1/2^(m-1) < 1/2^(N-1) < epsilon at the end? Shouldn't it remain <= ?

[theproofessayist]

I'm sorry so is the part where you show xn -x(n+1) is bounded by 2^-n is the same as x(n) - x(n+1) bounded by 2*epsilon just the working out the proof backwards to find N (even though that should not be the case since we're not given an explicit formula). I'm just wondering how the 2*epsilon helps if the real proof only starts at around 3:24.

[anjalibisht]

sir, how can we prove that whether or not the given Cauchy sequence is convergent?

[hugovera]

it is specified n>m, would we not need to show this still holds for m>n or m=n. So make it a proof with 3 cases unless there is a cool trick.

[olebogengmoeng]

under "advantage of using cauchy sequences" isn't that |Xm - Xn| <= |Xm - L| + |L - Xn| ?
And this is a really awesome video by the way.i also wanted to ask, do you have anything under THE BOLZANO-WEIERSTRASS topic of sequences?

[anjalibisht]

how to use Cauchy's general principle of convergence? plz help

[swat_katz_tbone]

I think there is a typo. The last term in the summation should be 1/(2^(n-1)) instead of 1/2^n

[MathDoctorBob]

Yes, that looks like a typo. I'll annotate. Thanks for the support. - Bob

[chandnibhudia]

im sorry isnt 1/2 + 1/4 + 1/8 + ie isnt it the sum of the sequence 1/2^n from n=1 to k as k approaches infinity? 

[MathDoctorBob]

@RespectMyHate Real analysis. Math majors have to go back and prove all the results from calculus rigorously. - Bob

[filcogaming]

my question is near the end, why did you multiply by 2 (at 5:39)