Proof that every convergent sequence is bounded

preview_player
Показать описание
A proof that every convergent sequence is bounded.
Рекомендации по теме
Комментарии
Автор

6 years later and still helping people! Thanks.

bbsilva
Автор

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!!!

danieltrigo
Автор

thank you so much that was so helpful, i was strugeling with this!! thanks from algeria <3

kiruvv
Автор

Thank you sir it is very helpful to me

dakshanakanapathy
Автор

Thank you very much! Got exam tomorrow, your video really helped! :)

snowyelsa
Автор

I was doing this and literally wrote down that maximum arguement for n<N but was having trouble realizing whether or not I was allowed to do that. So close, so very very close. Glad my ideas are validated. Thank you very much!

thedounut
Автор

Is Converse valid ??? Every bounded sequence is convergent ??? Here it is not monotonic

ananyasharma
Автор

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.

Pandafist
Автор

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?

cloudhuang
Автор

why is it still enough to consider only epsilon = 1?

gabrielmccartney
Автор

Why |a_n| <= max{|a_1, ..., a_N} for n<= N? Is there a specific reason that you chose n<= N? I know that it might be related to the definition of limit of a sequence but I don't understand how to picture it...

iOSGamingDynasties
Автор

Intuitively, why couldn't the terms of the sequence be infinity for some terms, then still approach some value when n approaches infinity?

EmapMe
Автор

why do we choose epsilon as 1? Can we choose any other number greater than 0?

monishadamodaran
Автор

How can |an| <= |aN| for all n<=N? E.g. an=1/n, for n<=N it follows that an > aN. What am I missing?

konohaninja
Автор

ur video is helpful but....it is not so clear to see...plz don't use black backgrounds

ShahidHussain-wjvx
Автор

Does this also apply for complex numbers? If not, how would the proof differ for them?

guydude
Автор

The conversly is true ?
we have a theorm say ( every bounded monotone sequence is convergant )

muhammedqadr
Автор

Every convergent sequence is bounded what about is converse? Prove both statements with examples

bilalwajid
Автор

How do we know |an|<=max{|a1|, ..., |aN|}... when I visualize it's obvious but I can't really prove it. Is this something from the definition of a convergent sequence or...?

popsicle
Автор

How do we know that an is less than or equal to max of a1....aN for n less than equal to N

noddyd