Detailed Proof of the Monotone Convergence Theorem | Real Analysis

[@WrathofMath]

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[@nullspace_xxii.]

your intuitive explanations are a force to be reckoned with!

[@WrathofMath]

I do my best - thanks for watching!

[@XahhaTheCrimson]

These days I review my undergraduate mathematics and this video is just perfect for that.... Thank you for clear explanation!

[@WrathofMath]

Glad to help, thanks for watching!

[@Joshua74801]

using this for my grad class. thanks

[@_kainya]

Thank you so much for this video, it sure helped me a lot!!! Your channel is highly underrated and your content is great! Your videos are really good!! I'm glad I found your channel!

[@WrathofMath]

Thank you!

[@vernitasutton4215]

Your videos help me understand the writings of my real analysis book. You should totally write a text!

[@WrathofMath]

Thanks so much, I'm glad to hear that! One day I'd love to write a text, though I wouldn't start with Real Analysis. Currently, my analysis playlist is mostly based on Real Analysis by Jay Cummings and Understanding Analysis by Abbott. I recommend both strongly!

[@johnvonneumannsdaddy8207]

tysm, my book proved the theorem in a really interesting way, but this clarifies it so much!

[@WrathofMath]

You're very welcome!

[@A7medK-wb5lk]

A great video! What drawing app did you use in this video?

[@oyeajugando7262]

Alabado seas!!! Te amoooo, me re ayudaste, Gracias!!!

[@WrathofMath]

¡De nada! ¡Muchas gracias por mirar!

[@lunkadapoorv985]

Why the assumption of monotonicity in monotone convergence theorem is necessary? Can you give me a detailed answer of this?

[@WrathofMath]

Thanks for watching and good question! Monotonicity is necessary because if a sequence is only bounded, it may not converge. Bounded sequences can still behave strangely - they can oscillate. Consider (-1)^n. This sequence is bounded below by -1 and above by 1, but it does not converge since it isn't monotone, it oscillates between 1 and -1. If a sequence is bounded, then it being monotone forces it to get closer and closer to its bound, and so we get convergence!

[@lunkadapoorv985]

@@WrathofMath ohh that was a wonderful explanation sir. Thank you so much, you made my day.

[@FullerHob]

We use the completeness property of the reals to conclude that sup(a n) exists when (a n) is increasing, but what do we use to conclude inf(a n) exists when the sequence is decreasing?

[@WrathofMath]

Thanks for watching and great question! We can use the completeness axiom to prove that a nonempty subset of the reals that is bounded below has an infimum. So I was taking that result for granted. It will be the next lesson I record, so give the proof a try yourself - it's fairly straightforward with some contradiction - and I'll reply with a link to the lesson when it is out!

[@FullerHob]

@@WrathofMath Thanks a lot for responding. Great video btw, you deserve a lot more views :-) I ended up working around it by taking a bounded increasing sequence that we just proved converges and timsing it by -1 so that it decreases, then concluding it converges to a limit which is the same as the increasing sequence’s limit timesed by -1 using the algebraic limit theorem

[@vedantsinha7374]

you're a life saver. Thank You

[@WrathofMath]

Glad to help, thanks for watching!

[@rayxxkaiser3586]

i don't understand for the summation of 1/n, where n is an integer from 1 to infinite. While ther series of 1/n to infinite decreases and tends to zero, the Integral test tell us the summation of 1/n is smaller than ln(n) - ln(1) = infinite. However, it fulfills the monotone convergence theoerm conditions, bounded and decreasing. Have I already confused these Mathematic ideas? I'm learning Integral and Series, so please help......many thanks

[@thomaspickin9376]

If you're looking at the infinite series 1/n (the summation), the sequence you're looking at is not (1, 1/2, 1/3, 1/4, 1/5, ... ), this just tells you what happens as n->inf in 1/n, not in a summation. This is monotone decreasing and bounded and so converges. The sequence you should be looking at is (1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, 1+1/2+1/4+1/5, ...), also known as the sequence of partial sums, this tells you what is happening in the sum. This is increasing and unbounded so diverges.

[@JTinfected]

Thank you soooo much sir🙏

[@antonincasadebaig2922]

Very helpful! thanks

[@WrathofMath]

Glad to hear it, thanks for watching! If you're looking for more real analysis, check out my analysis playlist and let me know if you have any video requests! kzbin.info/aero/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli

[@a.nelprober4971]

Bounded does not imply convergent surely? Consider (-1)^n

[@WrathofMath]

Correct, that is why the monotone convergence theorem also requires a function be monotone.

[@a.nelprober4971]

@@WrathofMath i would be so lost without your videos

[@WrathofMath]

Glad to help! Let me know if you have any video requests!

[@music_lyrics-ni7ks]

MONOTONE bounded. The sequence you mentioned, oscillates between 1 and -1, and hence cannot be monotone.

[@krasimirronkov17]

Didnt really understand why for increasing sequence the limit should be the supremum

[@bobek6155]

If the increasing sequence is unbounded -> diverges then the limit is infinity and so is the supremum. Bounded increasing sequence converges to something, that something is the limit therefore the limit is "the boundary" = supremum. Look up the definition of a supremum and it should become pretty clear.