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Math Help - Does this limit proof work?

  1. #1
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    Does this limit proof work?

    I have to prove that a_{n}=\frac {1+2+3+...+n}{n^{2}} converges to 1/2.

    Proof: We want to show that a_{n} converges to 1/2. So we want to show that
    \left|\frac {1+2+3+...+n}{n^{2}}-\frac{1}{2}\right|<\epsilon
    Note that a_{n}'=\frac {1+2+3+...+n}{n}>\frac {1+2+3+...+n}{n^{2}}=a_{n}.
    Thus if we prove the condition for a_{n}' then we simultaneously prove it for a_{n}.
    So a_{n}'-\frac{1}{2}=\frac{n(n+1)}{2n}-\frac{1}{2}=\frac{n(n+1)-n}{2n}=\frac{n}{2}.
    So take \frac{n}{2}<\epsilon \implies n<2\epsilon.
    So choose n^{*}<n<2\epsilon, and the condition holds for a_{n}' \implies the condition holds for a_{n} since a_{n}<a_{n}'. QED.

    Is this a valid proof?
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  2. #2
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    Why not use \displaystyle\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}?
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by paupsers View Post
    I have to prove that a_{n}=\frac {1+2+3+...+n}{n^{2}} converges to 1/2.

    Proof: We want to show that a_{n} converges to 1/2. So we want to show that
    \left|\frac {1+2+3+...+n}{n^{2}}-\frac{1}{2}\right|<\epsilon
    Note that a_{n}'=\frac {1+2+3+...+n}{n}>\frac {1+2+3+...+n}{n^{2}}=a_{n}.
    Thus if we prove the condition for a_{n}' then we simultaneously prove it for a_{n}. This is not true.
    So a_{n}'-\frac{1}{2}=\frac{n(n+1)}{2n}-\frac{1}{2}=\frac{n(n+1)-n}{2n}=\frac{n}{2}.
    So take \frac{n}{2}<\epsilon \implies n<2\epsilon. You want n to be greater than some function of epsilon, not less than it.
    So choose n^{*}<n<2\epsilon, and the condition holds for a_{n}' \implies the condition holds for a_{n} since a_{n}<a_{n}'. QED.

    Is this a valid proof?
    Proof:

    \left|\frac{\sum_{k=1}^n k}{n^2}-\frac{1}{2}\right|<\epsilon

    \left|\frac{\frac{n(n+1)}{2}}{n^2}-\frac{1}{2}\right|<\epsilon

    \left|\frac{n+1}{2n}-\frac{1}{2}\right|<\epsilon

    \left|\frac{n+1-n}{2n}\right|<\epsilon

    \left|\frac{1}{2n}\right|<\epsilon

    So for any \epsilon>0, let n>\frac{1}{2\epsilon}, and the condition will hold. \square
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