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Math Help - Another Sum Of This Series Problem

  1. #1
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    Another Sum Of This Series Problem

    Now in this problem I've attached, I was almost positive the answer would be 0 since the denominator will be approaching an infinitely large number, but I guess this isn't the case. Could someone tell me why the answer isn't 0 and how to go about doing this problem?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by fattydq View Post
    Now in this problem I've attached, I was almost positive the answer would be 0 since the denominator will be approaching an infinitely large number, but I guess this isn't the case. Could someone tell me why the answer isn't 0 and how to go about doing this problem?
    It's not zero since you're talking about a sum, not a limit. This problem is just a geometric series.

    \sum_{n=0}^{\infty}\left(\frac{1}{\sqrt{31}}\right  )^n = \frac{1}{1-\frac{1}{\sqrt{31}}} = -\frac{31}{\sqrt{31}-31}
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  3. #3
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    Hello, fattydq!

    \sum^{\infty}_{n=0}\frac{1}{(\sqrt{31})^n}

    Now in this problem I've attached, I was almost positive the answer would be 0 . no
    since the denominator will be approaching an infinitely large number.
    Be careful about what you're referring to . . .

    The terms are approach 0 . . . this is true.
    . . But the answer (what they are asking for) is the sum of the series.


    We have: . \sum^{\infty}_{n=0}\left(\frac{1}{\sqrt{31}}\right  )^n \;=\;1 + \frac{1}{\sqrt{31}} + \left(\frac{1}{\sqrt{31}}\right)^2 + \left(\frac{1}{\sqrt{31}}\right)^3 + \hdots

    This is a geometric series with first term a = 1 and common ratio r = \frac{1}{\sqrt{31}}

    Its sum is: . S \;=\;\frac{1}{1-\frac{1}{\sqrt{31}}} \;=\;\frac{\sqrt{31}}{\sqrt{31}-1} \;=\;\frac{\sqrt{31}(\sqrt{31}+1)}{30}

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