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Given $x_n$: $x_1 = 0$, $x_{2k} ={x_{2k - 1}}/2$, $\ x_{2k + 1} = 1 + x_{2k}$. Find $\liminf$ and $\limsup$, $n \to \infty$.

It seems obvious there only two subsequences and $\liminf x_n = 0$ for $n = 2k$ and $\limsup x_n = 1$ for $n = 2k + 1$. However I don't know how to find them without an explicit formula.

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  • $\begingroup$ I think it is possible to find explicit formuła for this sequence by inserting $x_{2k}$ into $x_{2k+1}$ and solving the recurrence equation with given initial conditions. $\endgroup$ Dec 17 '16 at 11:31
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We can derive separate formulas for $(x_{2k})$ and $(x_{2k+1})$.

For even indices we have: $x_2=0$ and $x_{2k+2}=\frac{1}{2}+\frac{x_{2k-2}}{2}$ for $k>1$. One can easily show by induction that this sequence is strictly increasing and bounded from above by $1$. Therefore, it has a limit, call it X. Then $X$ must satisfy $X=\frac{1}{2}+\frac{X}{2}$, which gives $X=1$.

For odd indices we have: $x_1=0$ and $x_{2k+1}=1+\frac{x_{2k-1}}{2}$. Again by induction, we see that this sequence is bounded from above by $2$ and is strictly increasing. Then it also converges to some limit, call it $Y$. Then $Y$ must satisfy $Y=1+Y/2$, which gives $Y=2$.

Since those are the only two converging subsequences, $\liminf=1$ and $\limsup=2$.

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