1. ## weird sequence

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

Show that $\displaystyle x_{n}$ does not converge.

Show that there is some subsequence of $\displaystyle x_{n}$ which converges to 0.

2. Originally Posted by frankdent1
Let a $\displaystyle \epsilon$ (0, 1) be an irrational number. Define a sequence $\displaystyle (x_{n})$ in [0, 1) by $\displaystyle x_{1}$ = 0 and
$\displaystyle x_{n+1}=\left\{\begin{array}{cc}x_{n}+a,&\mbox{ if } x_{n}+a<1\\x_{n}+a-1, &\mbox{ } otherwise \end{array}\right.$

Show that $\displaystyle x_{n}$ does not converge.

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

To find the subsequence, notice that $\displaystyle x_{n+1}$ is the fractional part of $\displaystyle x_n+a$, and therefore (by induction) $\displaystyle x_{n+1}$ is the fractional part of $\displaystyle na$. 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.