Superior and Inferior Limits of Sequences:

Question

Find the superior and inferior limits of the ${a_n}$ given that:

${a_n}$ defined by $a_1 = 0$, and for $k\in \mathbb{N}$,

$$ a_{2k}=\frac{1}{2}a_{2k-1},\ \ \ a_{2k+1}=\frac{1}{2}+a_{2k} $$

Answer 1

Look at two separate sequences: even and odd.

Odd: Look at the values that $a_1,a_3,a_5,a_7,...$ takes. Combine your two recurrence relations into one to get one that depends on odd indices only, $a_{2k+1} = \frac{1}{2}(1+a_{2k-1})$, which has the solution of $a_{2k+1}=1-2^{-k}$.

Even: look at $a_2,a_4,a_6,\ldots$. Combine the recurrence relations to get $a_{2k} = \frac{1}{2}\left(\frac{1}{2}+a_{2k-2}\right)$, with $a_2 = 0$. Here, we have that $a_2 = 0,a_4 = \frac{1}{4}, a_6 = \frac{3}{8},a_8 = \frac{7}{16}$, so $a_{2k} = \frac{2^{k-1}-1}{2^k}=\frac{1}{2}-2^{-k}$

From this, finding the sup and inf should be easy. You still have to prove that the closed forms I gave are correct, or give some other argument to that extent.