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Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0?
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Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1...
Introduction:
We are given a sequence {x(n)} such that the limit of the difference between consecutive terms of the sequence is a constant c. We need to determine the behavior of the sequence {x(n)/n} as n approaches infinity.
Claim: The sequence {x(n)/n} converges to zero.
Proof:
To prove this claim, we will use the definition of the limit of a sequence. Let ε > 0 be given. We need to find N such that for all n > N, |x(n)/n - 0| < />
Step 1: Find N1:
Since the limit of the difference between consecutive terms of {x(n)} is c, we have lim n->∞ (x(n+1) - x(n)) = c. Therefore, there exists N1 such that for all n > N1, |x(n+1) - x(n) - c| < />
Step 2: Find N2:
Next, we will find N2 such that for all n > N2, |x(n+1) - x(n) - c| < c.="" this="" can="" be="" done="" by="" choosing="" n2="" such="" that="" |x(n+1)="" -="" x(n)="" -="" c|="" />< c="" for="" all="" n="" /> N2.
Step 3: Find N:
Let N = max{N1, N2}. For all n > N, we have |x(n+1) - x(n) - c| < ε="" and="" |x(n+1)="" -="" x(n)="" -="" c|="" />< c.="" adding="" these="" two="" inequalities,="" we="" get="" 2|c|="" />< ε="" +="" />
Step 4: Final Proof:
Now, let's consider the term x(n)/n. We can write x(n)/n as (x(n+1) - x(n))/(n+1) - (x(n) - x(n-1))/n. Rearranging, we have x(n)/n = (x(n+1) - x(n))/(n+1) + (x(n-1) - x(n))/n.
Using the inequality |x(n+1) - x(n) - c| < ε="" and="" |x(n-1)="" -="" x(n)="" -="" c|="" />< ε="" (since="" n="" /> N), we can bound the terms in the above expression. We have |x(n)/n| < (ε="" +="" c)/(n+1)="" +="" />
Since ε + c and ε are constants, as n approaches infinity, both (ε + c)/(n+1) and ε/n approach zero. Therefore, the expression |x(n)/n| approaches zero as n approaches infinity.
Conclusion:
Thus, we have shown that for any ε > 0, there exists N such that for all n > N, |x(n)/n - 0| < ε.="" this="" satisfies="" the="" definition="" of="" the="" limit="" of="" a="" sequence,="" so="" the="" sequence="" {x(n)/n}="" converges="" to="" zero.="" therefore,="" the="" correct="" answer="" is="" option="" (d)="" -="" converges="" to="" 0.="" ε.="" this="" satisfies="" the="" definition="" of="" the="" limit="" of="" a="" sequence,="" so="" the="" sequence="" {x(n)/n}="" converges="" to="" zero.="" therefore,="" the="" correct="" answer="" is="" option="" (d)="" -="" converges="" to="" />
We are given a sequence {x(n)} such that the limit of the difference between consecutive terms of the sequence is a constant c. We need to determine the behavior of the sequence {x(n)/n} as n approaches infinity.
Claim: The sequence {x(n)/n} converges to zero.
Proof:
To prove this claim, we will use the definition of the limit of a sequence. Let ε > 0 be given. We need to find N such that for all n > N, |x(n)/n - 0| < />
Step 1: Find N1:
Since the limit of the difference between consecutive terms of {x(n)} is c, we have lim n->∞ (x(n+1) - x(n)) = c. Therefore, there exists N1 such that for all n > N1, |x(n+1) - x(n) - c| < />
Step 2: Find N2:
Next, we will find N2 such that for all n > N2, |x(n+1) - x(n) - c| < c.="" this="" can="" be="" done="" by="" choosing="" n2="" such="" that="" |x(n+1)="" -="" x(n)="" -="" c|="" />< c="" for="" all="" n="" /> N2.
Step 3: Find N:
Let N = max{N1, N2}. For all n > N, we have |x(n+1) - x(n) - c| < ε="" and="" |x(n+1)="" -="" x(n)="" -="" c|="" />< c.="" adding="" these="" two="" inequalities,="" we="" get="" 2|c|="" />< ε="" +="" />
Step 4: Final Proof:
Now, let's consider the term x(n)/n. We can write x(n)/n as (x(n+1) - x(n))/(n+1) - (x(n) - x(n-1))/n. Rearranging, we have x(n)/n = (x(n+1) - x(n))/(n+1) + (x(n-1) - x(n))/n.
Using the inequality |x(n+1) - x(n) - c| < ε="" and="" |x(n-1)="" -="" x(n)="" -="" c|="" />< ε="" (since="" n="" /> N), we can bound the terms in the above expression. We have |x(n)/n| < (ε="" +="" c)/(n+1)="" +="" />
Since ε + c and ε are constants, as n approaches infinity, both (ε + c)/(n+1) and ε/n approach zero. Therefore, the expression |x(n)/n| approaches zero as n approaches infinity.
Conclusion:
Thus, we have shown that for any ε > 0, there exists N such that for all n > N, |x(n)/n - 0| < ε.="" this="" satisfies="" the="" definition="" of="" the="" limit="" of="" a="" sequence,="" so="" the="" sequence="" {x(n)/n}="" converges="" to="" zero.="" therefore,="" the="" correct="" answer="" is="" option="" (d)="" -="" converges="" to="" 0.="" ε.="" this="" satisfies="" the="" definition="" of="" the="" limit="" of="" a="" sequence,="" so="" the="" sequence="" {x(n)/n}="" converges="" to="" zero.="" therefore,="" the="" correct="" answer="" is="" option="" (d)="" -="" converges="" to="" />
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Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0?
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Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0?.
Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let {x(n)} be a sequence or real numbers such that lim n-> inf (x (n 1)- x (n)) = c. Then the seq {x( n)/n} is a) is not bounded b) is bounded but not convergent c) converges to c d) converges to 0?.
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