a E R; when a = too, we leave the proof to the reader. Let Sn suplan, an+1,...}; so Sna. By the definition of the limit superior, for each k ≥ 1 there is an mk with a ≤ Smk a, then there are infinitely many values of n such that an> a. = (b) If a € R such that lim sup an a. Thus there is an N such that ank > a for all nk ≥ N. ank (b) If lim sup an

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter3: The Derivative
Section3.1: Limits
Problem 61E
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Need help with number 4, I have no idea how to do this problem. This is real analysis by the way.
a € R; when a = too, we leave the proof to the reader. Let Sn = suplan, an+1,...);
so Sna. By the definition of the limit superior, for each k ≥ 1 there is an
mk with a Smk <a + k-¹. By the definition of supremum there is an nk mk
such that a - k-1 < ang ≤ Smy. Thus la ang <k-1 and so the subsequence
{an} converges to a.
1.3.19. Corollary. Let {an} be a sequence in R.
(a) If a € R such that lim sup an> a, then there are infinitely many values of n such
that an> a.
(b) If a E R such that lim sup an <a, then there is an integer N such that an <a
for all n ≥ N.
Proof. (a) By the preceding proposition there is a subsequence {an} such that
ank→lim sup an> a. Thus there is an N such that an > a for all nk ≥ N.
(b) If lim sup an <a, then the definition of the lim sup implies there is an N with
sup{an, aN+1,...} <a. Hence part (b).
Exercises
(1) Prove Proposition 1.3.3(b).
(2)
For each of the following sequences {an} find the value of the limit and for each
stipulated value of e, find a value of N such that la - an <e when n ≥ N. (a) an =
n¹, e = .0001. (b) an=2", e = .0001. (Are the values for N you found the small-
est possible? This has no bearing on the convergence, but it's a bit more challenging
to find the smallest possible N.)
(3)
If ana and {an) is a renumbering of the original sequence, does any converge to
a?
(4) Prove that the sequence {n²} does not converge.
n+5
2n+6
(5) (a) Show that {2} converges and find its limit. (b) What is lim, ? (c) Show
that lim, 5n²+2
8n+10n²
12.
(6) Show the following. (a) √n+1-√n →0. (b) √n² + 2n-n→1. (c) √n+1
→ 0.
n
Transcribed Image Text:a € R; when a = too, we leave the proof to the reader. Let Sn = suplan, an+1,...); so Sna. By the definition of the limit superior, for each k ≥ 1 there is an mk with a Smk <a + k-¹. By the definition of supremum there is an nk mk such that a - k-1 < ang ≤ Smy. Thus la ang <k-1 and so the subsequence {an} converges to a. 1.3.19. Corollary. Let {an} be a sequence in R. (a) If a € R such that lim sup an> a, then there are infinitely many values of n such that an> a. (b) If a E R such that lim sup an <a, then there is an integer N such that an <a for all n ≥ N. Proof. (a) By the preceding proposition there is a subsequence {an} such that ank→lim sup an> a. Thus there is an N such that an > a for all nk ≥ N. (b) If lim sup an <a, then the definition of the lim sup implies there is an N with sup{an, aN+1,...} <a. Hence part (b). Exercises (1) Prove Proposition 1.3.3(b). (2) For each of the following sequences {an} find the value of the limit and for each stipulated value of e, find a value of N such that la - an <e when n ≥ N. (a) an = n¹, e = .0001. (b) an=2", e = .0001. (Are the values for N you found the small- est possible? This has no bearing on the convergence, but it's a bit more challenging to find the smallest possible N.) (3) If ana and {an) is a renumbering of the original sequence, does any converge to a? (4) Prove that the sequence {n²} does not converge. n+5 2n+6 (5) (a) Show that {2} converges and find its limit. (b) What is lim, ? (c) Show that lim, 5n²+2 8n+10n² 12. (6) Show the following. (a) √n+1-√n →0. (b) √n² + 2n-n→1. (c) √n+1 → 0. n
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