Find upper bound for sequence [duplicate]
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This question already has an answer here:
How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
5 answers
$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$
I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.
calculus sequences-and-series
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marked as duplicate by Martin R, Community♦ Feb 3 at 10:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
5 answers
$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$
I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.
calculus sequences-and-series
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marked as duplicate by Martin R, Community♦ Feb 3 at 10:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
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– Martin R
Feb 3 at 10:43
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Also related: math.stackexchange.com/q/525749/42969.
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– Martin R
Feb 3 at 10:44
add a comment |
$begingroup$
This question already has an answer here:
How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
5 answers
$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$
I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.
calculus sequences-and-series
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This question already has an answer here:
How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
5 answers
$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$
I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.
This question already has an answer here:
How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
5 answers
calculus sequences-and-series
calculus sequences-and-series
asked Feb 3 at 10:39
Nick KaklNick Kakl
176
176
marked as duplicate by Martin R, Community♦ Feb 3 at 10:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Community♦ Feb 3 at 10:46
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
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– Martin R
Feb 3 at 10:43
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Also related: math.stackexchange.com/q/525749/42969.
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– Martin R
Feb 3 at 10:44
add a comment |
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Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
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– Martin R
Feb 3 at 10:43
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Also related: math.stackexchange.com/q/525749/42969.
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– Martin R
Feb 3 at 10:44
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Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
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– Martin R
Feb 3 at 10:43
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Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
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– Martin R
Feb 3 at 10:43
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Also related: math.stackexchange.com/q/525749/42969.
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– Martin R
Feb 3 at 10:44
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Also related: math.stackexchange.com/q/525749/42969.
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– Martin R
Feb 3 at 10:44
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1 Answer
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Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.
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add a comment |
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Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.
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add a comment |
$begingroup$
Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.
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Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.
answered Feb 3 at 10:43
HarnakHarnak
1,3691512
1,3691512
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Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
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– Martin R
Feb 3 at 10:43
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Also related: math.stackexchange.com/q/525749/42969.
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– Martin R
Feb 3 at 10:44