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Math Help - [SOLVED] Geometric series and bounding

  1. #1
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    [SOLVED] Geometric series and bounding

    Find the values that bound (above and below) the geometric series (2/3)^n

    I don't even know where to start.
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  2. #2
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    Quote Originally Posted by Amberosia32 View Post
    Find the values that bound (above and below) the geometric series (2/3)^n

    I don't even know where to start.
    It depends on n.
    If n starts from zero, then \left(\frac{2}{3}\right)^0=1

    This is the largest value that \left(\frac{2}{3}\right)^n can be,
    since to form higher powers we are multiplying by a value <1 continuously.
    As n approaches infinity, the expression approaches zero.

    You should have an idea of what the graph of the curve looks like,
    as n ranges from 0 to infinity.

    If n is allowed to be negative, then

    \left(\frac{2}{3}\right)^{n} increase to infinity as n goes to minus infinity.

    This is because \frac{1}{\left(\frac{2}{3}\right)^{|n|}}=\left(\fr  ac{3}{2}\right)^{|n|}
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    It depends on n.
    If n starts from zero, then \left(\frac{2}{3}\right)^0=1

    This is the largest value that \left(\frac{2}{3}\right)^n can be,
    since to form higher powers we are multiplying by a value <1 continuously.
    As n approaches infinity, the expression approaches zero.

    You should have an idea of what the graph of the curve looks like,
    as n ranges from 0 to infinity.

    If n is allowed to be negative, then

    \left(\frac{2}{3}\right)^{n} increase to infinity as n goes to minus infinity.

    This is because \frac{1}{\left(\frac{2}{3}\right)^{|n|}}=\left(\fr  ac{3}{2}\right)^{|n|}
    ok, thanks
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  4. #4
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    Hi Amberosia32,

    you do need to clarify your question however.
    My answer only takes into account the graph of f(n)=\left(\frac{2}{3}\right)^n

    This may not be what you are looking for,
    as your post mentions "geometric series".

    If so, and \left(\frac{2}{3}\right)^n is not the formula for the sum,

    ie it may be a series of the form a+a\frac{2}{3}+a\left(\frac{2}{3}\right)^2+......

    If the series is something like this, then it needs to be evaluated first.

    Hence, do you have an entire question on this ?
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Hi Amberosia32,

    you do need to clarify your question however.
    My answer only takes into account the graph of f(n)=\left(\frac{2}{3}\right)^n

    This may not be what you are looking for,
    as your post mentions "geometric series".

    If so, and \left(\frac{2}{3}\right)^n is not the formula for the sum,

    ie it may be a series of the form a+a\frac{2}{3}+a\left(\frac{2}{3}\right)^2+......

    If the series is something like this, then it needs to be evaluated first.

    Hence, do you have an entire question on this ?
    I meant sequence. I'm sorry. And n start at 1, so it would be bounded between 2/3 and 0. I hope. lol.
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