Dude, I really appreciate your videos, they are a great contribution to my education. Thank you very much, and please keep going, your work is excellent!!!
Thanks for a clear and correct proof! Sequence proofs using the technique of max{} can be a little tricky and I've seen at least two professor incorrect solutions to this problem.
The Math Sorcerer thanks! Now i think i know it, and for the last part where we prove | a_n | N, this is based on definition of limit of sequence (there exists an N s.t. n>=N) and | a_n | < 1 + | L | right?
so if it converges, it's bounded, but if it is bounded, that doesn't mean it converges To recap, converges => bounded bounded and monotonic => converges
The result generalizes to any metric space (we must first write down the generalization of limit to metric spaces). The complex plane with the usual metric is one such space.
Excuse me sir, isn't there any possibility that a convergent sequence "blows up" divergently in some of the first "N" terms so that makes the sequence unbounded?
Because if you have a finite number of values you can take the maximum of the absolute value of those values, call it M. If it converges it is bounded by say a number N. Then take the max of these two and that’s a bound for your sequence.🔥
@@TheMathSorcerer for example, name a sequence being like this:1, 2, 3, infinity, 4, 4, 4, 4, 4......and so forth. Isn't this sequence convergent (converges to 4) but unbounded sir?
Thabk you. I found that your videos are very helpful. The explainations are clear n easy to follow.. except one thing.. the black background makes the text hard to see
Because a_n is actually an element of the sequence so it must be less than the max of that sequence or equal if a_n is itself the max of the sequence. Watch this for the detailed answer: ua-cam.com/video/0JtVbOohQUg/v-deo.html
@@aadhuu the question is about definitions of bounded. the answer is that the def of bounded in the video is equivalent to being bounded above and below
6 years later and still helping people! Thanks.
Wow it's been six years crazy!!
👍
@@TheMathSorcerer 10 years later aswell
Dude, I really appreciate your videos, they are a great contribution to my education. Thank you very much, and please keep going, your work is excellent!!!
I was doing this and literally wrote down that maximum arguement for n
The max of any finite set exists, so M, as defined in the video, exists.
Very sophisticated reasoning
thank you so much that was so helpful, i was strugeling with this!! thanks from algeria
Thank you very much! Got exam tomorrow, your video really helped! :)
:)
Thank you sir it is very helpful to me
Thanks for a clear and correct proof! Sequence proofs using the technique of max{} can be a little tricky and I've seen at least two professor incorrect solutions to this problem.
Nathan Woollard Np! Yes you have to be very careful!
Extremely clear! Thanks
glad it helped!!!!
Why |a_n|
it's true because it is(not a good answer I know lol). Maybe look at it like this: |a_n|
The Math Sorcerer thanks! Now i think i know it, and for the last part where we prove | a_n | N, this is based on definition of limit of sequence (there exists an N s.t. n>=N) and | a_n | < 1 + | L | right?
Yes! Strict inequality there at the end n >N but yes you got it!!!
@@TheMathSorcerer we say a_n≤ max {a1,a2....aN} because seq {an} may be increasing or decreasing right?
love from india❤
Is Converse valid ??? Every bounded sequence is convergent ??? Here it is not monotonic ......
There are bounded sequences that do not converge. For example a_n = sin(n), or a_n = (-1)^n, etc
so if it converges, it's bounded, but if it is bounded, that doesn't mean it converges
To recap,
converges => bounded
bounded and monotonic => converges
Yes! It helped me! Thank you :)
How can |an|
it helped a lot. thank you.
Does this also apply for complex numbers? If not, how would the proof differ for them?
The result generalizes to any metric space (we must first write down the generalization of limit to metric spaces). The complex plane with the usual metric is one such space.
why is it still enough to consider only epsilon = 1?
You can give any value for epsilon, it can be bigger or shorter and it does not matter or change anything.
why do we choose epsilon as 1? Can we choose any other number greater than 0?
ua-cam.com/video/MVbhdT0B7e4/v-deo.html
Watch this ..
Doubts will be absolutely cleared.
@@rishavmanhas NOOOOOOOOOOOOOOOOOOOOO IT'S PRIVAAAAATE :((((
Excuse me sir, isn't there any possibility that a convergent sequence "blows up" divergently in some of the first "N" terms so that makes the sequence unbounded?
No
Because if you have a finite number of values you can take the maximum of the absolute value of those values, call it M. If it converges it is bounded by say a number N. Then take the max of these two and that’s a bound for your sequence.🔥
@@TheMathSorcerer for example, name a sequence being like this:1, 2, 3, infinity, 4, 4, 4, 4, 4......and so forth. Isn't this sequence convergent (converges to 4) but unbounded sir?
Intuitively, why couldn't the terms of the sequence be infinity for some terms, then still approach some value when n approaches infinity?
How do we know |an|
well it's for n
Omg I don’t understand it😢
ur video is helpful but....it is not so clear to see...plz don't use black backgrounds
The conversly is true ?
we have a theorm say ( every bounded monotone sequence is convergant )
if it's bounded _and_ monotonic then it converges
there are plenty of bounded sequences that don't converge, for example a_n = (-1)^n, it's bounded since |a_n|
Thabk you. I found that your videos are very helpful. The explainations are clear n easy to follow.. except one thing.. the black background makes the text hard to see
Is its converse true or not?
No it's not take (-1) to the n it's bound Ed but diverges. Could not find the exponent key on my phone😄
How do we know that an is less than or equal to max of a1....aN for n less than equal to N
Because it’s a decreasing sequence?
Because a_n is actually an element of the sequence so it must be less than the max of that sequence or equal if a_n is itself the max of the sequence.
Watch this for the detailed answer: ua-cam.com/video/0JtVbOohQUg/v-deo.html
To be honest I'm still kind of confused. For the sequence to be bounded, does it not need to be bounded above and below?
Either I think.
by definition (a_n) is bounded by M if |a_n|=
@@rokettojanpu4669 if its an increasing sequence we can be sure the lower bound exists and vice versa..
i mean, if its a decreasing sequence, we can be sure the upper bound exists.
@@aadhuu the question is about definitions of bounded. the answer is that the def of bounded in the video is equivalent to being bounded above and below
Pls use yellow colour instead of red
Every convergent sequence is bounded what about is converse? Prove both statements with examples
Not every bounded sequence converges. Take the sequence (-1)^n.
Thank you!
I like your curly brackets. :)
Sir pls improve your writing..........m confused in words