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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?
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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.
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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?
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