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Math Help - Sum to infinity of a series

  1. #1
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    Sum to infinity of a series

    The first 3 terms of a series are given and the sum to n terms equation. Can someone explain what this technique does? It obviously divides each factor with n in it by n.
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  2. #2
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    Quote Originally Posted by Stuck Man View Post
    The first 3 terms of a series are given and the sum to n terms equation. Can someone explain what this technique does? It obviously divides each factor with n in it by n.
    For a constant C, \displaystyle \lim_{n\to\infty}\frac{C}{n}=0.

    By the way I assume that you cut off part of the problem statement at the top, where it should say something like "The formula S_n for the partial sums of the series"
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  3. #3
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    Hello, Stuck Man!

    \text{Given: the first 3 terms of a series and the sum to }n\text{ terms formula.}

    \text{Can someone explain what this technique does?}
    What is your question?
    Do you want their solution explained?


    \displaystyle}\frac{5}{2\cdot3\cdot4} + \frac{7}{3\cdot4\cdot5} + \frac{9}{4\cdot5\cdot6} + \hdots

    S_n \;=\;\dfrac{(2n+3)(2n+5)}{2(n+2)(n+3)} - \dfrac{5}{4}

    \text{Find the sum of the series to infinity.}


    \text{Write }S_n \text{ as: }\;\displaystyle \frac{(2+\frac{3}{n})(2+\frac{5}{n})}{2(1+\frac{2}  {n})(1+\frac{3}{n})} - \frac{4}{5}

    \text{Show that: }\;\displaystyle{\lim_{n\to\infty} S_n \;=\;\frac{3}{4}


    They gave us this formula: . S_n \;=\;\dfrac{(2n+3)(2n+5)}{2(n+2)(n+3)}


    Divide numerator and denominator by n^2\!:

    . . \displaystyle S_n \;=\; \frac{(\frac{2n+3}{n})(\frac{2n+5}{n})}{2(\frac{n+  2}{n})(\frac{n+3}{n})}-\frac{4}{5} \;=\; \frac{(2+\frac{3}{n})(2+\frac{5}{n})}{2(1+\frac{2}  {n})(1+\frac{3}{n})} - \frac{4}{5}


    Take the limit:

    . . \displaystyle \lim_{n\to\infty} S_n \;=\;\lim_{n\to\infty}\frac{(2+\frac{3}{n})(2+\fra  c{5}{n})}{2(1+\frac{2}{n})(1+\frac{3}{n})} - \frac{5}{4}

    . . . . . . . . \displaystyle =\; \frac{(2+0)(2+0)}{2(1+0)(1+0)} - \frac{5}{4} \;\;=\;\;\frac{4}{2} - \frac{5}{4} \;\;=\;\;\frac{3}{4}

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