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Math Help - Proof of a limit involving [x]

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    Super Member redsoxfan325's Avatar
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    Proof of a limit involving [x]

    Prove or disprove: \lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1, where [x] is the greatest integer function.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    Prove or disprove: \lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1, where [x] is the greatest integer function.
    This the same as \left[\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\righ  t]=\left[\frac{1}{1-\frac{1}{2}}-1\right]=\left[1\right]

    The greatest integer function of x is the same thing as the floor of x. Thus, \left[1\right]=\left\lfloor 1\right\rfloor=\boxed{1}
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    Quote Originally Posted by redsoxfan325 View Post
    Prove or disprove: \lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1, where [x] is the greatest integer function.
    for any positive integer n: \ \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 0, because 0 < \sum_{k=1}^n \frac{1}{2^k} < 1. so the limit is 0 not 1.
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    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    for any positive integer n: \ \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 0, because 0 < \sum_{k=1}^n \frac{1}{2^k} < 1. so the limit is 0 not 1.
    Yeah, you're definitely right, now that I think about it. Let \{x_n\} be a sequence where x_n = \left[\sum_{k=1}^n \frac{1}{2^k}\right]. In order for this sequence (and therefore the limit) to converge to 1, \forall~ \epsilon>0, \exists~ N such that \forall~ n>N, |1-x_n|<\epsilon, and this is clearly false, as \sum_{k=1}^n \frac{1}{2^k} < 1, so [x_n]=0 ~\forall~ n.
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    Quote Originally Posted by Chris L T521 View Post
    This the same as \left[\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\righ  t]=\left[\frac{1}{1-\frac{1}{2}}-1\right]=\left[1\right]

    The greatest integer function of x is the same thing as the floor of x. Thus, \left[1\right]=\left\lfloor 1\right\rfloor=\boxed{1}
    You can't put the limit inside the integer function.


    \sum_{k=1}^n \frac{1}{2^k}=\frac 12 \cdot \frac{1-\frac{1}{2^n}}{1-\frac 12}=1-\frac{1}{2^n}, which is obviously < 1.

    Putting the limit inside the integer function omits the fact that 1 is a limit, not the value of the sum.
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