The general term formula $a_{n+1}=\frac{1+a_n^2}2$ Let $\ a_1=\dfrac 12\ $ and $\ a_{n+1}=\dfrac{a_n^2+1}2$,


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*Could we find the general term formula $a_n$?

*If the answer to the question $1$ is "NO", for  $\left|a_n-\dfrac{n-1}{n+1}\right|$ and $\left|a_n-\dfrac{n}{n+1}\right|$, which one is small in comparison with the other?


The general term formula is difficult, as for the question $2$, $\left|a_n-\dfrac{n-1}{n+1}\right|> \left|a_n-\dfrac{n}{n+1}\right|$ ? I am not sure. I'd be grateful for any help you are able to provide.
 A: (1) As it was noted in the comments above, the sequence $(a_n/2)_{n\geq1}$ is defined by the iteration of the function $f(x)=x^2+c$ with $c=1/4$ and consequently it does not have a closed form.
(2) By a simple induction we see that $\frac{1}{2}\leq a_n<a_{n+1}<1$ for each $n$, This allows us to conclude that $(a_n)$ converges to the unique solution of the equation $x=(x^2+1)/2$ that is $1$.
Let us consider $b_n=1-a_n$. The sequence $(b_n)$ satisfies 
$$ b_1=\frac{1}{2},\quad b_{n+1}=b_n\left(1-\frac{b_n}{2}\right)$$
This recursion has a more convenient form:
$$
\frac{1}{b_{n+1}}-\frac{1}{b_n}=\frac{1}{2-b_n}\tag{1}
$$
 In particular, adding the inequalities $b_{k+1}^{-1}-b_{k}^{-1}\geq1/2$ for $k=1,\ldots,n-1$, we get
$b_n^{-1}\geq2+\frac{n-1}{2}$,
or equivalently
$$
b_n\leq \frac{2}{n+3}\tag{2}
$$
Using this again in $(1)$ we get $~b_{k+1}^{-1}-b_{k}^{-1}\leq\dfrac{1}{2}+\dfrac{1}{2(k+2)}$, adding these inequalities for $k=1,\ldots,n-1$, and rearranging we obtain
$b_n^{-1}\leq 2+\frac{n-1}{2}+\frac{1}{2}(H_{n+1}-3/2)$, with $H_n=\sum_{i=1}^n1/i$  (the well-known harmonic number.) Thus
$$
b_n\geq \frac{2}{n+3/2+H_{n+1}}\tag{3}
$$
Since clearly we have $H_{n+1}\leq n+\frac{1}{2}$ for $n\geq1$, and $n+3\geq n+1$ we conclude from $(1)$ and $(2)$ that
$$\frac{1}{n+1}\leq \frac{2}{n+3/2+H_{n+1}}\leq b_n\leq\frac{2}{n+3}\leq\frac{2}{n+1}$$
Now, note that
$$\eqalign{\left\vert a_n-\frac{n}{n+1}\right\vert-\left\vert a_n-\frac{n-1}{n+1}\right\vert&=
\left(b_n-\frac{1}{n+1}\right)-\left(\frac{2}{n+1}-b_n\right)
\cr&=2b_n-\frac{3}{n+1}\cr
&\geq  \frac{4}{n+3/2+H_{n+1}}-\frac{3}{n+1}\cr
&\geq\frac{2n-1-6H_{n+1}}{2(n+1)(n+3/2+H_{n+1})}
}
$$
Now, the sequence $(2n-1-6H_{n+1})$ is monotonous non-decreasing and the first positive term corresponds to $n=10$, (I checked this with Mathematica.) Thus, we have proved  that
$$
\forall\,n\geq 10,\quad
\left\vert a_n-\frac{n}{n+1}\right\vert>\left\vert a_n-\frac{n-1}{n+1}\right\vert
$$
For the first nine terms, we have to check the difference by "hand". Doing this yields the following conclusion:
$$
\forall\,n\geq 9,\quad
\left\vert a_n-\frac{n}{n+1}\right\vert>\left\vert a_n-\frac{n-1}{n+1}\right\vert
$$
and the inequality is reversed for $n\in\{1,2,\ldots,8\}$.
