Limit of positive definite sequence with terms $a_{n} = \frac{a_{n - 1} + 1}{2}$ or $a_{n} = sin(a_{n - 1})$ In the sequence all terms are positive. 
Each term can be either $a_{n} = \frac{a_{n - 1} + 1}{2}$ or $a_{n} = \sin{a_{n - 1}}$.
Can this sequence have a limit in the interval $(0, 1)$?
 A: No, it's not possible. Let us use an indirect prrof and assume that we have such a sequence, convergent to $g\in(0,1)$. Then by the definition of convergence $$ \forall \epsilon >0 \,\exists N \,\forall n>N :|a_n-g|<\epsilon$$
Let us take $\epsilon < \min\{\frac{1-g}{4},\frac{g-\sin g}{2}\}$. For $n$ big enough we have $|a_n-g|<\epsilon$ and $|a_{n+1}-g|<\epsilon$. We have two cases.
If $a_{n+1} = \frac{a_n+1}{2}$, then
$$ |a_{n+1}-g| = |\frac{a_n+1}{2} - g| = |\frac{a_n-g}{2} + \frac{1-g}{2}| \ge \big| |\frac{1-g}{2}| - |\frac{a_n-g}{2}|\big| \rightarrow^{n\rightarrow\infty} \frac{1-g}{2}$$
so for $n$ big enough $|a_{n+1}-g| > \frac{1-g}{4} \ge \epsilon$.
If $a_{n+1} = \sin a_n$, then
$$ |a_{n+1}-g| = |\sin a_n - g| \ge \big| |g-\sin g| - |\sin a_n - \sin g|\big| \rightarrow^{n\rightarrow\infty} g-\sin g $$
so for $n$ big enough $|a_{n+1}-g| > \frac{g-\sin g}{2} \ge \epsilon$.
We get a contradiction with the assumption that $|a_{n+1}-g|<\epsilon$.
In conclusion it's impossible to get a convergent series; if one element of the sequence gets close to the would-be limit, then the next one cannot be arbitratily close.
A: For the sake of simplicity, suppose $a_{2n+1}=\frac{a_{2n} + 1}{2}$ and $a_{2n+2}=\sin(a_{2n+1})$, with $0<a_1$.
If $\lim a_n=L$ exists, we have $L=\frac{L+1}{2}$ and $L=\sin(L)$, i.e., $\sin(1)=1$, which is absurd.
So, the sequence can converge only if there finitely many terms of one of the forms.
