# Convergence (or not) of some sequence

Let an arbitrary sequence $(a_{n})_{n=1}^{\infty}$, where all $a_{n}\in(0,1)$, be given. Is the sequence $(b_{n})_{n=1}^{\infty}$, where $b_{1}=a_{1}$, $b_{n+1}=(1-b_{n})+(2b_{n}-1)a_{n}$, convergent? Numerical computations suggest it's always convergent and usually its limit is $\frac{1}{2}$. Other limits can be available for instance in the case when $a_{n}$ tends to 1. No idea, however, how to proof mathematically.
Marc

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What happens to the sequence where $a_n$ is: $0.9, 0.99, 0.999, \ldots$? – Benjamin Dickman Jan 28 '13 at 9:55
Numerical it convergences, but its limit clearly depends on the initial term of $b_{n}$ (but I can't see no relationship between them, though). Marc – user59934 Jan 28 '13 at 10:14

If $(a_n)$ does not converge to $1$, then $b_n$ in fact converges to $1/2$.
Proof. Let $c_n=b_n-1/2$. Then $$c_1=a_1-1/2,\quad c_{n+1}=(2\,a_n-1)c_n$$ and $$|c_{n+1}|=|2\,a_n-1|\,|c_n|<|c_n|.$$ The sequence $(|c_n|)$ is decreasing and bounded below by $0$, so that it converges to some $c\ge0$. If $c\ne0$ then $$|2\,a_n-1|=\frac{|c_{n+1}|}{|c_n|}$$ converges to $1$. This in turn implies that $(a_n)$ converges to $1$, arriving at a contradiction.
If $\lim_{n\to\infty}a_n=1$, then other limits are possible. Let $\epsilon_n=1-a_n$. Then $\epsilon_n\ge0$ and $\lim_{n\to\infty}\epsilon_n=0$, $$b_{n+1}=b_n-\epsilon_n(2\,b_n-1)$$ and $$|b_{n+1}-b_n|\le\epsilon_n.$$ If $\sum_{n=1}^\infty\epsilon_n<\infty$, then $(b_n)$ converges.
The only case left is when $\sum_{n=1}^\infty\epsilon_n=\infty$.