Let $\ a_1=\dfrac 12\ $ and $\ a_{n+1}=\dfrac{a_n^2+1}2$,
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.
(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 $n\in\{1,2,\ldots,8\}$.
$$b_n = b_{n-1},^2 + c$$
In this case use $b_n = \frac 12 a_n$ which gives $b_1 = \frac 14$ and $c = \frac 14$. The sequence starts at $1 \over 4$ and grows from below to $1 \over 2$.
– DanielV Apr 11 '14 at 16:44