Set $b_{n}= \frac{1}{a_{n}}$. Then $b_{n}$ obeys the recurrence \[b_{n+1}= b_{n}-\frac{1}{2006+1/b_{n}}\, .\] In particular, we get the inequality \[b_{n+1}> b_{n}-\frac{1}{2006}\, .\] Iterating that inequality shows that $b_{n}> 2-\frac{n}{2006}$, which is at least $1$ for $n \le 2006$. In particular, we have $a_{2006}< 1$. For $n \le 2006$, we have shown $b_{n}> 1$. Plugging back in the original recurrence for $b_{n}$, for $n \le 2006$ we get the inequality \[b_{n+1}< b_{n}-\frac{1}{2007}\, .\] By iterating that inequality, we see that $b_{2006}< 2-\frac{2006}{2007}= \frac{2008}{2007}$. Taking reciprocals gives $a_{2006}> \frac{2007}{2008}= 1-\frac{1}{2008}$.

shobber wrote: $a_{n+1}=a_{n}+\frac{1}{2006}a_{n}^{2}$ , $n \in N$, $a_{0}=\frac{1}{2}$. Prove that $1-\frac{1}{2008}< a_{2006}< 1$. This is a very interesting problem. First of all it reduces to iteration of $f(x) = x+x^{2}$, starting from $x = 1/4012$ (divide the given equation by 2006). Next, for small $x$, $f(x)$ will be very close to $x+x^{2}+x^{3}+x^{4}+...$ and iteration of that series can be carried out explicitly; it sends $1/n$ to $1/(n-1)$. The problem is saying that if we start from $x = 1/2n$ then this appoximation is highly accurate even for $n$ steps of the iteration. It cannot be accurate forever, because eventually the values must become large.