From ${a_{n + 1}} = \frac{{{a_n}({a_n} + 2012)}}{{2012}}$ we have $\frac{1}{{{a_{n + 1}}}} = \frac{1}{{{a_n}}} - \frac{1}{{{a_n} + 2012}}$, so $\frac{1}{{{a_n} + 2012}} = \frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}$.Beacause ${a_k} < 1 < {a_{k + 1}}$, $\sum\limits_{i = 0}^k {\frac{1}{{{a_i} + 2012}}} = \frac{1}{{{a_0}}} - \frac{1}{{{a_{k + 1}}}} > 1$, and $\sum\limits_{i = 0}^{k - 1} {\frac{1}{{{a_i} + 2012}}} = \frac{1}{{{a_0}}} - \frac{1}{{{a_k}}} < 1$。 For $\sum\limits_{i = 0}^{2011} {\frac{1}{{{a_i} + 2012}}} < \frac{{2012}}{{{a_0} + 2012}} < 1$, we have $k \ge 2012$, so ${a_{2012}} < 1$. Hence$\sum\limits_{i = 0}^{2012} {\frac{1}{{{a_i} + 2012}}} > \frac{{2013}}{{{a_{2012}} + 2012}} > 1$, so $k \le 2012$, and we have $k = 2012$.

（2）： prove that: $\displaystyle \lim_{n\to\infty}\frac{n}{lnn}({na_{n}+2012})=2012^2 $

tian_275461 wrote: （2）： prove that: $\displaystyle \lim_{n\to\infty}\frac{n}{lnn}({na_{n}+2012})=2012^2 $ Are you sure? $ \lim_{n\to\infty}\frac{n}{\ln n}=\infty$ ?