4. Let Sn cos(7). Use the definition to find lim sup så and lim inf sn, then explain why s has no limit. (hint Thm 10.7) =
4. Let Sn cos(7). Use the definition to find lim sup så and lim inf sn, then explain why s has no limit. (hint Thm 10.7) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Transcribed Image Text:Sn,
4. Let Sn
cos(2). Use the definition to find lim supsn and lim inf
3
then explain why sn has no limit. (hint Thm 10.7)
=
![10.7 Theorem.
Let (sn) be a sequence in R.
(i) If lim sʼn is defined [as a real number, +∞ or -], then
lim inf
lim Sn
=
lim sup Sn.
lim sup sn, then lim sn is defined and lim sn
= lim sup Sn.
Sn
(ii) If lim inf sn
lim inf Sn
-
-
Proof
We use the notation uy = inf{sn : n > N}, vN
U = lim un
lim inf sn and v =
lim UN
= lim sup Sn.
(i) Suppose lim sn = +∞o. Let M be a positive real number. Then
there is a positive integer N so that
n > N implies
Sn > M.
Then UN = inf{sn n > N} ≥ M. It follows that m > N
implies um ≥ M. In other words, the sequence (un) satisfies
the condition defining lim uy = +∞o, i.e., lim inf sn = +∞.
Likewise lim supsn = +∞.
The case lim sn = ∞ is handled in a similar manner.
Now suppose lim sn = s where s is a real number. Consider
€ > 0. There exists a positive integer N such that |sn − s| < €
for n > N. Thus Sn < s + e for n > N, so
=
sup{sn n > N},
UN = sup{sn: n > N} ≤s+e.
=
Also, m > N implies um ≤s+€, so lim sup sn - lim vm ≤ s+€.
Since lim sup sn ≤ s + e for all € > 0, no matter how small, we
conclude lim sup sn ≤ s lim sn. A similar argument shows
lim slim inf sn. Since lim inf sn ≤ lim sup sn, we infer all
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d5e88e6-b1af-4aea-9b08-2dadd85f5e2c%2Ffecbfdca-eed7-4844-b9b6-1c3d0fc63aa7%2Frfeacq8_processed.png&w=3840&q=75)
Transcribed Image Text:10.7 Theorem.
Let (sn) be a sequence in R.
(i) If lim sʼn is defined [as a real number, +∞ or -], then
lim inf
lim Sn
=
lim sup Sn.
lim sup sn, then lim sn is defined and lim sn
= lim sup Sn.
Sn
(ii) If lim inf sn
lim inf Sn
-
-
Proof
We use the notation uy = inf{sn : n > N}, vN
U = lim un
lim inf sn and v =
lim UN
= lim sup Sn.
(i) Suppose lim sn = +∞o. Let M be a positive real number. Then
there is a positive integer N so that
n > N implies
Sn > M.
Then UN = inf{sn n > N} ≥ M. It follows that m > N
implies um ≥ M. In other words, the sequence (un) satisfies
the condition defining lim uy = +∞o, i.e., lim inf sn = +∞.
Likewise lim supsn = +∞.
The case lim sn = ∞ is handled in a similar manner.
Now suppose lim sn = s where s is a real number. Consider
€ > 0. There exists a positive integer N such that |sn − s| < €
for n > N. Thus Sn < s + e for n > N, so
=
sup{sn n > N},
UN = sup{sn: n > N} ≤s+e.
=
Also, m > N implies um ≤s+€, so lim sup sn - lim vm ≤ s+€.
Since lim sup sn ≤ s + e for all € > 0, no matter how small, we
conclude lim sup sn ≤ s lim sn. A similar argument shows
lim slim inf sn. Since lim inf sn ≤ lim sup sn, we infer all
=
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