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  1. #1
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    weird sequence

    Let a \epsilon (0, 1) be an irrational number. Define a sequence (x_{n}) in [0, 1) by x_{1} = 0 and
    x_{n+1}=\left\{\begin{array}{cc}x_{n}+a,&\mbox{ if }<br />
x_{n}+a<1\\x_{n}+a-1, &\mbox{ } otherwise \end{array}\right.

    Show that x_{n} does not converge.

    Show that there is some subsequence of x_{n} which converges to 0.
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  2. #2
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    Quote Originally Posted by frankdent1 View Post
    Let a \epsilon (0, 1) be an irrational number. Define a sequence (x_{n}) in [0, 1) by x_{1} = 0 and
    x_{n+1}=\left\{\begin{array}{cc}x_{n}+a,&\mbox{ if }<br />
x_{n}+a<1\\x_{n}+a-1, &\mbox{ } otherwise \end{array}\right.

    Show that x_{n} does not converge.

    Show that there is some subsequence of x_{n} which converges to 0.
    If x_n\to \ell then \ell must be equal to either \ell+a or \ell+a-1. You can easily see that neither of these cases is possible.

    To find the subsequence, notice that x_{n+1} is the fractional part of x_n+a, and therefore (by induction) x_{n+1} is the fractional part of na. It will be sufficient to show that we can find multiples of a with fractional parts arbitrarily close to 0.

    Divide the unit interval into k subintervals of length 1/k. The fractional parts of na are all distinct (otherwise a would be rational), and at least one of the subintervals must contain infinitely many of them. So we can find arbitrarily large m and n (with n>m) for which the fractional parts of ma and na differ by less than 1/k. Then the fractional part of (n-m)a is either less than 1/k, or it is greater than 1-(1/k) in which case the fractional parts of r(n-m)a (for r=1,2,3,...) will move down through the unit interval in steps of less than 1/k, until one of them is less than 1/k.
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