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Consider the seqn {sn} where sn = sin(nπθ), where θ be a rational no. such that 0 < θ < 1, then,
- a)(sn) be convergent seqn
- b)(sn) is not a convergent seqn
- c)(sn) has a subsequence which convergent to 0.
- d)(sn) has a subsequence which may or may not converge.
Correct answer is option 'B,C,D'. Can you explain this answer?
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Verified Answer
Consider the seqn {sn} where sn = sin(nπθ), whereθbe a ...
[Note that 0 < p/q <1]. Thus a contains a subsequence whose limit is 0 and a subsequence which (may or may not converge but certainly) does not have the limit zero. Hence, the given sequence is not convergent.
Most Upvoted Answer
Consider the seqn {sn} where sn = sin(nπθ), whereθbe a ...
The sequence {sn} = sin(n) is a sequence of sine values where each term is obtained by taking the sine of the corresponding natural number.
For example, the first few terms of the sequence would be:
s1 = sin(1) = 0.841
s2 = sin(2) = 0.909
s3 = sin(3) = 0.141
s4 = sin(4) = -0.757
s5 = sin(5) = -0.959
...
The values of the sine function range from -1 to 1, so each term in the sequence {sn} will also be between -1 and 1.
Since the sine function is periodic with a period of 2π, the sequence {sn} will repeat itself after every 2π terms. This means that the sequence will never converge to a single value and will oscillate indefinitely between -1 and 1.
In summary, the sequence {sn} where sn = sin(n) is an oscillating sequence that takes on values between -1 and 1.
For example, the first few terms of the sequence would be:
s1 = sin(1) = 0.841
s2 = sin(2) = 0.909
s3 = sin(3) = 0.141
s4 = sin(4) = -0.757
s5 = sin(5) = -0.959
...
The values of the sine function range from -1 to 1, so each term in the sequence {sn} will also be between -1 and 1.
Since the sine function is periodic with a period of 2π, the sequence {sn} will repeat itself after every 2π terms. This means that the sequence will never converge to a single value and will oscillate indefinitely between -1 and 1.
In summary, the sequence {sn} where sn = sin(n) is an oscillating sequence that takes on values between -1 and 1.
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Consider the seqn {sn} where sn = sin(nπθ), whereθbe a ...
[Note that 0 < p/q <1]. Thus a contains a subsequence whose limit is 0 and a subsequence which (may or may not converge but certainly) does not have the limit zero. Hence, the given sequence is not convergent.
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Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer?
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Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer?.
Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer?.
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Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer?, a detailed solution for Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? has been provided alongside types of Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Consider the seqn {sn} where sn = sin(nπθ), whereθbe a rational no. such that 0 <θ< 1, then,a)(sn) be convergent seqnb)(sn) is not a convergent seqnc)(sn) has a subsequence which convergent to 0.d)(sn) has a subsequence which may or may not converge.Correct answer is option 'B,C,D'. Can you explain this answer? tests, examples and also practice Mathematics tests.
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