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Math Help - Series - nth term - Is this right?

  1. #1
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    Series - nth term - Is this right?

    I am not 100% confident about this so would someone mind checking my work?

    The nth partial sum of a Series is given by:
    Sn=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\  frac{1}{2n}=\sum_{k=1}^{n}\frac{1}{n+k}

    1. What is the associated Series:

    \sum_{n=1}^{\infty}\frac{1}{2n}

    2. Convergence?

    \lim_{n\to\infty}\frac{1}{2n}=0

    General term goes to 0

    Test for convergence by direct comparison.
    \frac{1}{2n}=\frac{1}{n-\frac{1}{2}}>\frac{1}{n}

    \frac{1}{n} divergent harmonic series

    \sum_{n=1}^{\infty}\frac{1}{2n} diverges by direct comparison with \frac{1}{n}

    3. Determine the Limit if possible

    \lim_{n\to\infty}\;Sn=\lim_{n\to\infty}\frac{1}{2n  }=0
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  2. #2
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    Quote Originally Posted by tiredNConfused View Post
    I am not 100% confident about this so would someone mind checking my work?

    The nth partial sum of a Series is given by:
    Sn=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\  frac{1}{2n}=\sum_{k=1}^{n}\frac{1}{n+k}

    1. What is the associated Series:
    Let a_n = \left\{ \begin{array}{c} \frac{1}{2} \text{ for }n=1 \\ \frac{1}{2n-1}+\frac{1}{2n} \text{ for }n\geq 2 \end{array} \right.

    Then the series is \sum_{n=1}^{\infty} a_n

    This series converges if and only if \sum_{n=2}^{\infty} a_n = \sum_{n=2}^{\infty} \left( \frac{1}{2n-1} + \frac{1}{2n} \right) converges.

    Can you finish this now?
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  3. #3
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    Yes I can, I knew it wasn't as easy as I made it.

    But one question, how did you come up with:  \frac{1}{2n-1}
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  4. #4
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    Quote Originally Posted by tiredNConfused View Post
    Yes I can, I knew it wasn't as easy as I made it.

    But one question, how did you come up with:  \frac{1}{2n-1}
    Because,
    s_1 = \boxed{a_1 = \tfrac{1}{2}}
    s_2 = \boxed{a_1+a_2 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}}
    s_3 = \boxed{a_1+a_2+a_3 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5  }+\tfrac{1}{6}}

    Now solve the first boxed equation, well it is already solved, a_1 = \tfrac{1}{2}.

    For the second boxed equation we have a_1 + a_2 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4} and we know what a_1 is.
    Solving it gives, a_2 = \tfrac{1}{3}+\tfrac{1}{4}.

    Similar with the third boxed equation we get a_3 = \tfrac{1}{5} + \tfrac{1}{6}

    And this generalizes. That is how we get the formula for a_1.
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  5. #5
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    Thanks soooo much!!!
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