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    thanks
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    I don't think it has a special name.... Just a series.

    Why? Are you trying to sum it?
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    (Original post by Daniel Freedman)
    I don't think it has a special name.... Just a series.

    Why? Are you trying to sum it?
    ye working out convergence dont know how.
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    (Original post by ibbi786)
    ye working out convergence dont know how.
    What tests do you know that tell you about convergence?
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    (Original post by Daniel Freedman)
    What tests do you know that tell you about convergence?
    not to sure was trying ratio test but got lost
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    (Original post by ibbi786)
    not to sure was trying ratio test but got lost
    The ratio test states that, if you have  \displaystyle \sum_{n=1}^{\infty} a_n , then:

    - If  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 , the sum converges

    - If  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1 , the sum diverges

    - If  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1 , then it doesn't tell us anything.

    Have you tried calculating   \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ? Where did you get stuck?
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    (Original post by Daniel Freedman)
    The ratio test states that, if you have  \displaystyle \sum_{n=1}^{\infty} a_n , then:

    - If  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 , the sum converges

    - If  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1 , the sum diverges

    - If  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1 , then it doesn't tell us anything.

    Have you tried calculating   \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ? Where did you get stuck?
    ive found out it converges but dont know how to calculate what it converges to...
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    (Original post by ibbi786)
    ive found out it converges but dont know how to calculate what it converges to...
    Ok. I'll show you how to do it, but I'll spoiler sections appropriately to give you the opportunity to benefit from thinking about it. Please do think about it.

    There's a nice trick to summing series like these that is useful to learn. The idea is to start off considering

     (1-x)^{-1} = 1 + x + x^2 + x^3 + ...  \ \ \ (*)

    and then fiddle with it, to make the RHS look like our series.

    (you can see this is true, as RHS is an infinite geometric series with common ratio x)

    Now, if we substitute x = 1/2 into both sides, we get something that looks promising:

     (1-\frac{1}{2})^{-1} = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + ...

    It has the increasing powers of 1/2 that we need, but it doesn't have the right numbers on top (i.e. the 1, 2, 3 etc in our original series). Can you think of something we can do to (*), to get these numbers to appear?

    Spoiler:
    Show
    Differentiate both sides.

     (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + ... \ \ \ (**)

    Now, let's see what happens if you plug in x = 1/2:

     (1-\frac{1}{2})^{-2} = 1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + ... ,

    which is almost right, but the powers of two in the denominator of each term is one too low. What could we do to (**) to sort this out?

    Spoiler:
    Show
    Multiply through by x.

     x(1-x)^{-2} = x + 2x^2 + 3x^3 + 4x^4 + ...

    Now when put put x = 1/2, we get our original series on the RHS. So we can deduce that the series converges to:

     \frac{1}{2}(1-\frac{1}{2})^{-2} = 2


    [Note, all these series are nice because x = 1/2 lies nicely within their radii of convergence]
 
 
 
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