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Math Help - limit

  1. #1
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    limit

    I'm doing my homework and got stuck on these problems. Can someone please help. Thank you much.


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  2. #2
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    Hi.

    1) Do you have to use the hint? Alternatively there is the infinite geometric series

    s_n := \sum_{k=1}^n \frac{1}{2^k}


    j:=k-1


    s_n := \sum_{j=0}^{n-1} \frac{1}{2^{j+1}} = \sum_{j=0}^{n-1} \frac{1}{2^{j}}*2^{-1} = 1/2 * \sum_{j=0}^{n-1} \frac{1}{2^{j}}


    Let n \to \infty


    0.5* \sum_{j=0}^{\infty} \frac{1}{2^j} = 0.5 * 2 = 1
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by john_n82 View Post
    I'm doing my homework and got stuck on these problems. Can someone please help. Thank you much.


    s_n-\frac{1}{2}s_n = \sum_{k=1}^n \frac{1}{2^k} - \sum_{k=1}^n \frac{1}{2^{k+1}} = \sum_{k=1}^n \frac{1}{2^{k+1}} = \sum_{k=2}^{n+1} \frac{1}{2^k} = \sum_{k=1}^{n+1}\left(\frac{1}{2^k}\right)-\frac{1}{2} = s_{n+1}-\frac{1}{2}

    The rest of the solution to 1 lies below if you need it.
    Spoiler:
    Thus, \frac{1}{2}s_n = s_n-\frac{1}{2}s_n = s_{n+1}-\frac{1}{2}.

    Letting \lim_{n\to\infty} s_n = s, we have \frac{1}{2}s = s-\frac{1}{2}, implying that s=1.
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  4. #4
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    Oh, so for #1, I don't have to use the definition of the limit (no epsilon)???
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  5. #5
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    Quote Originally Posted by john_n82 View Post
    Oh, so for #1, I don't have to use the definition of the limit (no epsilon)???
    That's right, there is no need to use any epsilon
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  6. #6
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    are there any hints on #2, 3, 4? Thank you much.
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