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Radius of convergence of the power series Watch

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    I have got to the point where I have

    | k(1-x) |
    -----------
    (k+1)

    Where exactly do I go from here to find the radius of convergence?
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    (Original post by Bruce Harrisface)
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    I have got to the point where I have

    | k(1-x) |
    -----------
    (k+1)

    Where exactly do I go from here to find the radius of convergence?

    I am a bit busy to do the question but assuming what you are writing is correct, then you want the limit of this as k tends to infinity to be strictly less than 1

    This will give you the range of x

    (PS If this is what year 9s do these days, then the Government reforms in education are clearly working)
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    (Original post by Bruce Harrisface)
    Name:  ImageUploadedByStudent Room1453374653.315312.jpg
Views: 55
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    I have got to the point where I have

    | k(1-x) |
    -----------
    (k+1)

    Where exactly do I go from here to find the radius of convergence?
    By the ratio test, a series with terms u_n converges (absolutely) if \lim_{n \to \infty} |\frac{u_{n+1}}{u_n}| < 1 and diverges if the same limit is > 1.

    In your case, you are considering \lim_{n \to \infty} |\frac{(x-1)^{n+1}/n+1}{(x-1)^n/n}| = |\frac{n(x-1)}{n+1}| - note that the -1 terms mod to 1, so I haven't shown them.

    Bear in mind that x is some arbitrary real number. So your job is to:

    1. find the limit of the expression above - the x term doesn't take part in the limiting process though - it's just some arbitrary number.
    2. solve an inequality imposed by the ratio test to find the set of values of x which satisfy it.

    In addition, the ratio test says nothing if the limit *is* 1 - you will have to consider these cases separately to see if the end points 1 and -1 also give convergence.
 
 
 
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