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Math Help - sequence question

  1. #1
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    sequence question

    How do I interpret \frac{1}{2^{(n-1)}+1}+\frac{1}{2^{(n-1)}+2}+\frac{1}{2^{(n-1)}+3}+...+\frac{1}{2^n}
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  2. #2
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    2^{n-1}+k advances from k=1 up to k=2^{n-1}, at which point 2^{n-1}+2^{n-1}=2^n
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  3. #3
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    Quote Originally Posted by manyarrows View Post
    How do I interpret \frac{1}{2^{(n-1)}+1}+\frac{1}{2^{(n-1)}+2}+\frac{1}{2^{(n-1)}+3}+...+\frac{1}{2^n}
    The sum of the recipricals of all the natural numbers greater than 2^{n-1} and less than or equal to 2^n

    CB
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  4. #4
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    Quote Originally Posted by manyarrows View Post
    How do I interpret \frac{1}{2^{(n-1)}+1}+\frac{1}{2^{(n-1)}+2}+\frac{1}{2^{(n-1)}+3}+...+\frac{1}{2^n}
    What do you mean by "interpret"? Were the instructions something like "describe the rule for the n-th term in words"...?

    Thank you!
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  5. #5
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    No it was to verify the inequality with that series > or = to 1/2.
    How do I copy an equation from a previous thread and edit it in a new thread? Anyway , I have problem with sequences or series when the last term differs from the first. I don't know they look. For example in the above how does n=3 look compared to n=5? How do you incorporate the last term.
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  6. #6
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    \sum_{k=1}^{2^{n-1}} \frac{1}{2^{n-1} + k} \geq \sum_{k=1}^{2^{n-1}} \frac{1}{2^{n-1} + 2^{n-1}}=\frac{1}{2^n} \sum_{k=1}^{2^{n-1}} 1 = \frac{2^{n-1}}{2^n}=\frac{1}{2}.
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