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sequences question

Hi

If b_n>=a_n

and also a_n tends to infinity

how do you show b_n tends to infinity, thanks
(edited 5 years ago)
Original post by joker55
Hi

If b_n>=a_n

and also a_n tends to infinity

how do you show a_n tends to infinity, thanks
Please correct the statement.
Reply 2
Original post by ftfy
Please correct the statement.

corrected
Original post by joker55
corrected

What's the definition for a sequence tending to infinity? (You should be able to use this and literally one line of reasoning to answer the question).
Reply 4
Original post by DFranklin
What's the definition for a sequence tending to infinity? (You should be able to use this and literally one line of reasoning to answer the question).

So you use the definition of infinity for a_n and somehow use the information that b_n>=a_n to show b_n tends to infinity?
Original post by joker55
So you use the definition of infinity for a_n and somehow use the information that b_n>=a_n to show b_n tends to infinity?

Yes. (Without being snarky, what on earth else could you possibly do?)
Reply 6
Original post by DFranklin
Yes. (Without being snarky, what on earth else could you possibly do?)

thanks. what does b_n>=a_n mean, the reason i cant answer it is because i dont know what this means
Original post by joker55
thanks. what does b_n>=a_n mean, the reason i cant answer it is because i dont know what this means

It's the "plain-text" way of writing bnanb_n \geq a_n.
Original post by joker55
thanks. what does b_n>=a_n mean, the reason i cant answer it is because i dont know what this means


It's a bit surprising you're asked to prove that statement without being told what it means. Then again, you could just blindly apply it in the definition and end up with the proof.

It just means that if we denote

{an}={a1,a2,a3,} \{ a_n \} = \{a_1, a_2, a_3, \ldots \}

and

{bn}={b1,b2,b3,}\{ b_n \} = \{ b_1, b_2, b_3, \ldots \}

and so bnanb_n \geq a_n means b1a1b_1 \geq a_1 and b2a2b_2 \geq a_2 and b3a3b_3 \geq a_3 and so on...

So every term of the sequence {bn}\{ b_n \} is greater than or equal to the respective element of the sequence {an}\{ a_n \}
(edited 5 years ago)
ana_n \to \infty means for every M>0M > 0, there exists an NNN \in \mathbb{N} such that an>Ma_n > M for all n>Nn > N.

Now you're told that bnanb_n \ge a_n. Do you see it?
Reply 10
Original post by DFranklin
It's the "plain-text" way of writing bnanb_n \geq a_n.






so all the terms in b_n are greater than or equal to those of a_n, so if

Original post by RDKGames
It's a bit surprising you're asked to prove that statement without knowing this.

It just means that if we denote

{an}={a1,a2,a3,} \{ a_n \} = \{a_1, a_2, a_3, \ldots \}

and

{bn}={b1,b2,b3,}\{ b_n \} = \{ b_1, b_2, b_3, \ldots \}

and so bnanb_n \geq a_n means b1a1b_1 \geq a_1 and b2a2b_2 \geq a_2 and b3a3b_3 \geq a_3 and so on...

So every term of the sequence {bn}\{ b_n \} is greater than or equal to the respective element of the sequence {an}\{ a_n \}

is it that A sequence (an) tends to infinity if, for every C > 0, there exists

N N such that a_n > C whenever n > N so b_n tends to infinity?

and as b_n>a_n>C for all n in N b_n tends to infinity
(edited 5 years ago)
Original post by joker55
so all the terms in b_n are greater than or equal to those of a_n, so if


is it that A sequence (an) tends to infinity if, for every C > 0, there exists

N N such that b_n>a_n > C whenever n > N so b_n tends to infinity?


Yes, precisely that.

In the definition for (an)(a_n) \to \infty you have the requirement that an>Ca_n > C. But since bnanb_n \geq a_n then bnan>Cb_n \geq a_n > C hence bn>Cb_n > C.

So now you just have bnb_n satisfying the definition hence it tends to infinity.
Original post by joker55
so all the terms in b_n are greater than or equal to those of a_n, so if


is it that A sequence (an) tends to infinity if, for every C > 0, there exists

N N such that a_n > C whenever n > N so b_n tends to infinity?

and as b_n>a_n>C for all n in N b_n tends to infinity

I would not be happy with what you've written, although I suspect you've got the right idea.

The flipside of "this proof has only one line of actual reasoning" is that if you don't give that line of reasoning in a clear way, it's hard to give much credit for what you have done.

Your proof/explanation should look like:

"By definition of ana_n \to \infty, {definition goes here}".

But we know that bnanb_n \geq a_n

Therefore {actual reasoning step}, and so {something looking very like your first definition}

So we deduce bnb_n \to \infty".
Reply 13
Original post by DFranklin
I would not be happy with what you've written, although I suspect you've got the right idea.

The flipside of "this proof has only one line of actual reasoning" is that if you don't give that line of reasoning in a clear way, it's hard to give much credit for what you have done.

Your proof/explanation should look like:

"By definition of ana_n \to \infty, {definition goes here}".

But we know that bnanb_n \geq a_n

Therefore {actual reasoning step}, and so {something looking very like your first definition}

So we deduce bnb_n \to \infty".

Thanks for you help

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