<s>[INST] Finding the value of $\prod_{n=0}^\infty a_n$ with $a_0=1/2$ and $a_{n}=1+(a_{n-1}-1)^2$ 
Putting the value of the first term, we can see that the series goes like
$$1/2, 5/4, 17/16,...$$ 
I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
 [/INST] Let $b_n=a_n-1$ so that $b_0=-\frac 12$ and $b_{n}=b_{n-1}^2$.
Consequently, $b_n=b_0^{2^n}$ hence
$$a_n=1+(-2)^{-2^n}$$
Let $x=\frac 12$, then
\begin{align}
(1+x)\prod_{n=0}^Na_n
&=(1+x)\prod_{n=0}^N(1+(-x)^{2^n})\\
&=(1+x)(1-x)(1+x^2)\cdots(1+x^{2^N})\\
&=(1-x^2)(1+x^2)\cdots(1+x^{2^N})\\
&=(1-x^4)(1+x^4)\cdots(1+x^{2^N})\\
&=(1-x^8)(1+x^8)\cdots(1+x^{2^N})\\
&=\cdots\\
&=(1-x^{2^N})(1+x^{2^N})\\
&=1-x^{2^{N+1}}\\
&\xrightarrow{N\to\infty}1
\end{align}
so that
$$\prod_{n=0}^\infty a_n=\frac 23$$
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