Mathematics Exam > Mathematics Questions > Let {an}, {bn} and {cn} be sequences of real ...
Start Learning for Free
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergent
- a)implies both {bn} and {cn} are convergent
- b)implies {bn} is convergent but {cn} need not be convergent
- c)implies {cn} is convergent but {bn} need not be convergent
- d)if both {bn} and {cn} are convergent
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2...
If {an} is convergent,
Then {bn} and {cn} both are convergent. because {bn} and {cn} are sub sequence of {an}
Then {bn} and {cn} both are convergent. because {bn} and {cn} are sub sequence of {an}
Most Upvoted Answer
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2...
Explanation:
To prove that option A is the correct answer, we need to show that if the sequence {an} is convergent, then both {bn} and {cn} are also convergent.
Convergence of a Sequence:
A sequence {an} is said to be convergent if there exists a real number L such that for every positive number ε, there exists a positive integer N such that for all n ≥ N, |an - L| < />
To prove that {an} implies both {bn} and {cn} are convergent, we need to show that for every ε > 0, there exists a positive integer N1 and N2 such that for all n ≥ N1, |bn - L1| < ε="" and="" for="" all="" n="" ≥="" n2,="" |cn="" -="" l2|="" />< />
Proof:
Let's assume that {an} is convergent. Then, there exists a real number L such that for every ε > 0, there exists a positive integer N such that for all n ≥ N, |an - L| < />
Convergence of {bn}:
Now, let's consider the sequence {bn} = {a2n}. We need to show that {bn} is convergent. Let's choose an arbitrary ε > 0.
Since {an} is convergent, there exists a positive integer N1 such that for all n ≥ N1, |an - L| < />
Let's substitute n = 2k in the above inequality, where k is a positive integer.
|a2k - L| < />
Now, let's consider the difference between bn and L:
|bn - L| = |a2k - L|
Since n = 2k, we can rewrite the above inequality as:
|bn - L| < />
Therefore, for every ε > 0, there exists a positive integer N1 such that for all n ≥ N1, |bn - L| < ε.="" this="" shows="" that="" {bn}="" is="" />
Convergence of {cn}:
Now, let's consider the sequence {cn} = {a2n+1}. We need to show that {cn} is convergent. Let's choose an arbitrary ε > 0.
Since {an} is convergent, there exists a positive integer N2 such that for all n ≥ N2, |an - L| < />
Let's substitute n = 2k + 1 in the above inequality, where k is a positive integer.
|a2k+1 - L| < />
Now, let's consider the difference between cn and L:
|cn - L| = |a2k+1 - L|
Since n = 2k + 1, we can rewrite the above inequality as:
|cn - L| < />
Therefore, for every ε > 0, there exists a positive integer N2 such that for all n ≥ N2, |cn - L| < ε.="" this="" shows="" that="" {cn}="" is="" />
Conclusion:
Since we have shown that if {an} is convergent, then both {bn} and {cn} are convergent, the correct answer is option A.
To prove that option A is the correct answer, we need to show that if the sequence {an} is convergent, then both {bn} and {cn} are also convergent.
Convergence of a Sequence:
A sequence {an} is said to be convergent if there exists a real number L such that for every positive number ε, there exists a positive integer N such that for all n ≥ N, |an - L| < />
To prove that {an} implies both {bn} and {cn} are convergent, we need to show that for every ε > 0, there exists a positive integer N1 and N2 such that for all n ≥ N1, |bn - L1| < ε="" and="" for="" all="" n="" ≥="" n2,="" |cn="" -="" l2|="" />< />
Proof:
Let's assume that {an} is convergent. Then, there exists a real number L such that for every ε > 0, there exists a positive integer N such that for all n ≥ N, |an - L| < />
Convergence of {bn}:
Now, let's consider the sequence {bn} = {a2n}. We need to show that {bn} is convergent. Let's choose an arbitrary ε > 0.
Since {an} is convergent, there exists a positive integer N1 such that for all n ≥ N1, |an - L| < />
Let's substitute n = 2k in the above inequality, where k is a positive integer.
|a2k - L| < />
Now, let's consider the difference between bn and L:
|bn - L| = |a2k - L|
Since n = 2k, we can rewrite the above inequality as:
|bn - L| < />
Therefore, for every ε > 0, there exists a positive integer N1 such that for all n ≥ N1, |bn - L| < ε.="" this="" shows="" that="" {bn}="" is="" />
Convergence of {cn}:
Now, let's consider the sequence {cn} = {a2n+1}. We need to show that {cn} is convergent. Let's choose an arbitrary ε > 0.
Since {an} is convergent, there exists a positive integer N2 such that for all n ≥ N2, |an - L| < />
Let's substitute n = 2k + 1 in the above inequality, where k is a positive integer.
|a2k+1 - L| < />
Now, let's consider the difference between cn and L:
|cn - L| = |a2k+1 - L|
Since n = 2k + 1, we can rewrite the above inequality as:
|cn - L| < />
Therefore, for every ε > 0, there exists a positive integer N2 such that for all n ≥ N2, |cn - L| < ε.="" this="" shows="" that="" {cn}="" is="" />
Conclusion:
Since we have shown that if {an} is convergent, then both {bn} and {cn} are convergent, the correct answer is option A.
|
Explore Courses for Mathematics exam
|
|
Similar Mathematics Doubts
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer?
Question Description
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. 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 Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer?.
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. 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 Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer?.
Solutions for Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.
Here you can find the meaning of Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Let {an}, {bn} and {cn} be sequences of real numbers such that bn = a2n and cn = a2n+1. Then {an} is convergenta)implies both {bn} and {cn} are convergentb)implies {bn} is convergent but {cn} need not be convergentc)implies {cn} is convergent but {bn} need not be convergentd)if both {bn} and {cn} are convergentCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mathematics tests.
|
|
Explore Courses for Mathematics exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup