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Math Help - Infinite Series

  1. #1
    Member rtblue's Avatar
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    Infinite Series

    Evaluate:

    \frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{3}{16}+\  frac{5}{32}...

    I know the formula for an infinite series:

    \frac{a}{1-r}


    But i cannot seem to find the rate. I have tried grouping the terms, and doing two infinite series, but I still cannot find a pattern. Any help is appreciated.
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  2. #2
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    Hello, rtblue!

    Is there a typo?
    Shouldn't the third fraction be \frac{1}{8}\,?


    \displaystyle \text{Evaluate: }\:S \:=\:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{3}{  16}+\frac{5}{32} + \hdots

    We have: . \displaystyle S \;=\;\left(\frac{1}{2} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{3}{16} + \frac{5}{32} + \frac{7}{64} + \hdots\right)

    . . . . ,. . . \displaystyle S \;=\;\frac{3}{8} + \frac{1}{8}\underbrace{\left(1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \hdots\right)}_{\text{This is }T} .[1]


    \begin{array}{ccccc}<br />
\text{We have:} & T &=& 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + \hdots & [2]\\ \\[-3mm]<br />
\text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}T &=& \quad\;\; \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \hdots & [3]\end{array}

    Subtract [2]-[3]: . \frac{1}{2}T \;=\;\;1 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \hdots

    . . . . . . . . . . . . . \frac{1}{2}T \;=\; 1 + \underbrace{\left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \hdots\right)}_{\text{This sums to 2}}

    . . . . . . . . . . . . . \frac{1}{2}T \;=\;1 + 2 \quad\Rightarrow\quad T \:=\:6


    Substitute into [1]: . \displaystyle S \;=\;\frac{3}{8} + \frac{1}{8}(6) \;=\;\frac{9}{8}

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  3. #3
    Member rtblue's Avatar
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    I do not believe there is a typo, but I am taking the question from an online source, so there might be some error there. Anyways, the answer given to my question is 2. Also, i have noticed that the numerators form the fibonacci series. I am not sure if this helps though.
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  4. #4
    Member rtblue's Avatar
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    Actually i have just found a formula, for a problem like this, involving the fibonacci series:

    \frac{r}{r^2-(r+1)}

    for:

    \frac{1}{r}+\frac{1}{r^2}+\frac{2}{r^3}+\frac{3}{r  ^4}...

    and the numerators form the Fibonacci series.

    note that the second 1/4 can be expressed as 2/8.

    after substituting r, we get the sum to be 2.
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