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Math Help - limit at infinity of sum of arithmetic sequence

  1. #1
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    limit at infinity of sum of arithmetic sequence

    First,
    a = (1 + 2 + 3 + ... + n) / (n^2)

    When I try to find the limit of a when n tends to infinity,

    I have tried two ways:

    1. I converted a to
    2 - [(1 / 2) ^ (n) ] and get the value 2.

    2. I make it to

    a = 1 / n^2 + 2 / n^2 + 3 / n^2 + ... + n/ n^2 and then take the limit.
    but this give an answer 0.

    Why? I know there must be something wrong.
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  2. #2
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    Quote Originally Posted by ling_c_0202 View Post
    First,
    a = (1 + 2 + 3 + ... + n) / (n^2)

    When I try to find the limit of a when n tends to infinity,

    I have tried two ways:

    1. I converted a to
    2 - [(1 / 2) ^ (n) ] and get the value 2.

    2. I make it to

    a = 1 / n^2 + 2 / n^2 + 3 / n^2 + ... + n/ n^2 and then take the limit.
    but this give an answer 0.

    Why? I know there must be something wrong.
    Hello,

    use the property:

    1 + 2 + 3 + ... + n = 1/2*n*(n+1)

    Using the usual methods you'll get 1/2 as the limit if n approaches infinity.

    EB
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,

    use the property:

    1 + 2 + 3 + ... + n = 1/2*n*(n+1)

    Using the usual methods you'll get 1/2 as the limit if n approaches infinity.

    EB


    But what's wrong with my second method?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ling_c_0202 View Post
    But what's wrong with my second method?
    The sum 1/n^2 + ... n/n^2 is an infinite sum. You can't just use the limit of the last term to find the limit of the sum.

    -Dan
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  5. #5
    Grand Panjandrum
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    \lim_{n \to \infty} \frac{\sum_{r=1}^n r}{n^2}=\lim_{n \to \infty}\frac{n(n+1)/2}{n^2} =\lim_{n \to \infty} \left[n^2/(2n^2)+n/(2n^2)\right]=\lim_{n \to \infty}\left[1/2+1/(2n)\right]=1/2
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