1. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$.
Then the integer part of $\sum_{n=0}^{201} \frac{1}{a_{n}+1}$ is $\qquad$.
(2011, National High School Mathematics League Gansu Province Preliminary)
Given: $\frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}}$
$$
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\frac{1}{a_{0}}-\frac{1}{a_{2012}}=4-\frac{1}{a_{2012}} \text {. }
$$

Obviously, the sequence $\left\{a_{n}\right\}$ is monotonically increasing.
$$
\begin{array}{l}
\text { Also, } a_{1}=\frac{5}{16}, a_{2}>\frac{5}{16}+\frac{5}{16} \times \frac{1}{4}=\frac{25}{64}, \\
a_{3}>\frac{25}{64}+\frac{25}{64} \times \frac{1}{3}=\frac{25}{48}>\frac{1}{2}, \\
a_{4}>\frac{1}{2}+\frac{1}{4}=\frac{3}{4}, a_{5}=\frac{3}{4}+\frac{9}{16}>1,
\end{array}
$$

Thus, $a_{2012}>1$.
Therefore, $0<\frac{1}{a_{2012}}<1, \sum_{n=0}^{2011} \frac{1}{a_{n}+1} \in(3,4)$.
Hence, the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is 3.