2. The sequence $\left(a_{n}\right)_{n=1}^{\infty}$ is defined by $a_{1}=\frac{1}{2}, a_{n+1}=a_{n}-a_{n}^{2}$. Is $a_{999}<\frac{1}{1000}$?
Solution. I method. From $a_{n+1}=a_{n}-a_{n}^{2}$, we obtain

$$
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(1-a_{n}\right)}=\frac{1}{a_{n}}+\frac{1}{1-a_{n}}
$$

From this,

$$
\frac{1}{a_{999}}=\frac{1}{a_{998}}+\frac{1}{1-a_{998}}=\frac{1}{a_{997}}+\frac{1}{1-a_{997}}+\frac{1}{1-a_{998}}=\ldots=\frac{1}{a_{1}}+\frac{1}{1-a_{1}}+\frac{1}{1-a_{2}}+\ldots+\frac{1}{1-a_{998}}
$$

Given that $0 < a_{n} < 1$, it follows that $\frac{1}{1-a_{n}} > 1$. Therefore,

$$
\frac{1}{a_{999}}=\frac{1}{a_{1}}+\sum_{n=1}^{998} \frac{1}{1-a_{n}}>2+998=1000
$$

From this, $a_{999} < \frac{1}{1000}$.

---

Solution. II method. Let $b_{n}=\frac{1}{a_{n}}$. Then, from $a_{n+1}=a_{n}-a_{n}^{2}$, we get

$$
b_{n+1}=\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(1-a_{n}\right)}=\frac{1}{a_{n}}+\frac{1}{1-a_{n}}=b_{n}+\frac{1}{1-\frac{1}{b_{n}}}=b_{n}+\frac{b_{n}}{b_{n}-1}=b_{n}+1+\frac{1}{b_{n}-1}
$$

From this, $b_{n}>b_{1}+(n-1)=n+1$, so $b_{999}>1000$, hence $a_{999}<\frac{1}{1000}$.