<s>[INST] Consider the recursively defined sequence $x_{1} =0$ $x_{2n} =x_{2n-1}/2$ and $x_{2n+1}=(1/2) + x_{2n}$ I've posted this question before, but there was a typo which messed up my search for a solution
$x_{1} =0$, $x_{2n} =x_{2n-1}/2$ and $x_{2n+1}=1/2 + x_{2n}$
1)Which is the limit of the sequences $x_{2n}$ and $x_{2n+1}$?
I thought I could substitute $$\lim_{n\rightarrow\infty}x_{2n-1} = \lim_{n\rightarrow\infty}x_{2n+1}$$ but then I got stuck and quite don't really know what to do
2)Does $x_{n}$ converges? If yes to which point?
Thanks in advance.
 [/INST] First, you can substitute $x_{2n}$ in $x_{2n+1} = \frac{1}{2} + x_{2n}$ with $x_{2n} = \frac{1}{2} x_{2n-1} $ to get $x_{2n+1} = \frac{1}{2} + \frac{1}{2}x_{2n-1}$.
Define another sequence $y_n = x_{2n+1}$, so $y_0 = x_1 = 0$ and $y_n = x_{2n+1} = \frac{1}{2} + \frac{1}{2} x_{2(n-1) + 1} = \frac{1}{2} + \frac{1}{2}y_{n-1}$. Solution of these reccurent equation is $y_n = 1 - \frac{1}{2^n}$, so $x_{2n+1} = 1 - \frac{1}{2^n}$.
From $x_{2n} = \frac{1}{2} x_{2n-1}$ it's now obvious that $x_{2n} = \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2^{n-1}} = \frac{1}{2} - \frac{1}{2^n}$, so
$\lim_{n\to\infty} x_{2n} = \frac{1}{2}$ and $\lim_{n\to\infty} x_{2n+1} = 1$. As these limits aren't equal, the sequence $x_n$ does not converge.
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