- #1
hiro
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I have two questions involving subsequential limits. One I have started and understand what I need to do, the second I really don't know how to start.
First question
Let 0 ≤ a < b < +∞. Define the sequence a[itex]_{n}[/itex] recursively by setting a[itex]_{1}[/itex] = a, a[itex]_{2}[/itex] = b, and a[itex]_{n+2}[/itex]=(a[itex]_{n}[/itex]+a[itex]_{n+1}[/itex])/2 [itex]\forall[/itex]n.
Show that the sequences a[itex]_{2n}[/itex] and a[itex]_{2n-1}[/itex] are monotonic and convergent. Does a[itex]_{n}[/itex] converge? To what?
Solution attempt
This is the second part of the question. The first part I already proved, and shows that if a[itex]_{2n}[/itex] and a[itex]_{2n-1}[/itex] both converge to the same limit then a[itex]_{n}[/itex] converges to that limit.
I can show that both subsequences are bounded. I can also show that if their limits exists, they are the same (by plugging in a[itex]_{2n}[/itex]→L a[itex]_{2n-1}[/itex]→L and using the recurrence relations). The limit is obviously a+(2/3)b (though I need to prove it).
So all I need to do is show that a[itex]_{2n-1}[/itex] is nondecreasing and a[itex]_{2n}[/itex] is nonincreasing. Unfortunately I can't find the right induction proof to show this. Then I need to find one of the limits (either the full sequence or the even or odd subsequence).
Second question
Let λ[itex]\in[/itex][0,1]. Show that there exists a sequence r[itex]_{n}[/itex] such that r[itex]_{n}[/itex]→λ. r[itex]_{n}[/itex][itex]\in[/itex]{0,1/2[itex]^n[/itex],2/2[itex]^n[/itex],...,(2[itex]^n[/itex]-1)/2[itex]^n[/itex],1}.
Solution attempt
The correct theorem is presumably:
Let S denote the set of subsequential limits of a sequence s[itex]_{n}[/itex]. Suppose t[itex]_{n}[/itex] is a sequence in S[itex]\cap[/itex]ℝ and that t[itex]_{n}[/itex]→t. Then t belongs to S.
However I can't figure out how to apply it.
Thanks for your help.
First question
Let 0 ≤ a < b < +∞. Define the sequence a[itex]_{n}[/itex] recursively by setting a[itex]_{1}[/itex] = a, a[itex]_{2}[/itex] = b, and a[itex]_{n+2}[/itex]=(a[itex]_{n}[/itex]+a[itex]_{n+1}[/itex])/2 [itex]\forall[/itex]n.
Show that the sequences a[itex]_{2n}[/itex] and a[itex]_{2n-1}[/itex] are monotonic and convergent. Does a[itex]_{n}[/itex] converge? To what?
Solution attempt
This is the second part of the question. The first part I already proved, and shows that if a[itex]_{2n}[/itex] and a[itex]_{2n-1}[/itex] both converge to the same limit then a[itex]_{n}[/itex] converges to that limit.
I can show that both subsequences are bounded. I can also show that if their limits exists, they are the same (by plugging in a[itex]_{2n}[/itex]→L a[itex]_{2n-1}[/itex]→L and using the recurrence relations). The limit is obviously a+(2/3)b (though I need to prove it).
So all I need to do is show that a[itex]_{2n-1}[/itex] is nondecreasing and a[itex]_{2n}[/itex] is nonincreasing. Unfortunately I can't find the right induction proof to show this. Then I need to find one of the limits (either the full sequence or the even or odd subsequence).
Second question
Let λ[itex]\in[/itex][0,1]. Show that there exists a sequence r[itex]_{n}[/itex] such that r[itex]_{n}[/itex]→λ. r[itex]_{n}[/itex][itex]\in[/itex]{0,1/2[itex]^n[/itex],2/2[itex]^n[/itex],...,(2[itex]^n[/itex]-1)/2[itex]^n[/itex],1}.
Solution attempt
The correct theorem is presumably:
Let S denote the set of subsequential limits of a sequence s[itex]_{n}[/itex]. Suppose t[itex]_{n}[/itex] is a sequence in S[itex]\cap[/itex]ℝ and that t[itex]_{n}[/itex]→t. Then t belongs to S.
However I can't figure out how to apply it.
Thanks for your help.