Prove that a sequence which is bounded above cannot tend to infinity

[garyljc]

question 1 : Prove that a sequence which is bounded above cannot tend to infinity
What i did was state the definition ... but I'm trying to proof by contradiction. So i first suppose that a(n) tends to infinity , then a(n) > C . But since it is bounded above , C < or = to U , where U is the upper bound .
This is where i got stuck. Any ideas ?

question 2 : I am required to prove the this sequence does not tend to infinity
B(n) = cos(n^2 + 7)
how am i suppose to do this ?

 

[HallsofIvy]

question 1 : Prove that a sequence which is bounded above cannot tend to infinity
What i did was state the definition ... but I'm trying to proof by contradiction. So i first suppose that a(n) tends to infinity , then a(n) > C . But since it is bounded above , C < or = to U , where U is the upper bound .
This is where i got stuck. Any ideas ?

Saying "a(n)> C" makes no sense until you have said what C is!

question 2 : I am required to prove the this sequence does not tend to infinity
B(n) = cos(n^2 + 7)
how am i suppose to do this ?

Well, just a random guess, but since you were just asked to prove that "if a sequence is bounded above is cannot tend to infinity", how about finding an upper bound for cos(n²+ 7)? There's an obvious one.

 

[garyljc]

i got question 2
does is this statement true : that for any bounded sequence, it cannot tend to infinity ?

for question 1 , a sequence tend to infinity for every C>0 , there exist a nat no. N such that a(n) > C whenever n>N
what do you mean i have not said what is C ?
isn't c just a number ?

 

[HallsofIvy]

i got question 2
does is this statement true : that for any bounded sequence, it cannot tend to infinity ?

That is exactly what you were asked to prove in question 1!

for question 1 , a sequence tend to infinity for every C>0 , there exist a nat no. N such that a(n) > C whenever n>N
what do you mean i have not said what is C ?
isn't c just a number ?

You statement here, "a sequence tends to infinity for every C> 0" makes no sense (the sequence does not depend on any C) and that is NOT what you said in your first post: "Prove that a sequence which is bounded above cannot tend to infinity"- there is no "C" in that.

My point about not saying what C was is that you said earlier: "So i first suppose that a(n) tends to infinity , then a(n) > C" and THAT makes no sense because you haven't said what C is. You did not say, for example that C was a number. Even if you did say "C is a number" you still wouldn't have proven it. Yes, you must have $C\le U$, but if C were, for example, 2, it might be that the upperbound on {a(n)} was 3! What you meant to say, and should say, is that if {a(n)} "tends to infinity", then for C any number, there exist N such that a(N)> C. If the upperbound on {a(n)} is U, take C= U+ 1. Then what do you have?

 

[garyljc]

but it reads here from my notes that the definition is stated as i said ... could i then have a correct definition of a sequence that tends to infinity ?

 

[HallsofIvy]

What you wrote, and what I was objecting to was:
" a sequence tend to infinity for every C>0 "
which should be " a sequence tend to infinity IF for every C>0".