2. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}>1, a_{n+1}=a_{n}+\frac{1}{a_{n}^{2}} \text {. }
$$

Then the sequence $\left\{a_{n}\right\}()$.
(A) is bounded and decreasing
(B) is bounded and increasing
(C) is unbounded and increasing
(D) is unbounded and decreasing
2.C.

From $a_{n+1}=a_{n}+\frac{1}{a_{n}^{2}}>a_{n}$, we know that $\left\{a_{n}\right\}$ is increasing.
Notice that
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}^{2}} \\
\Leftrightarrow a_{n+1}^{3}=\left(a_{n}+\frac{1}{a_{n}^{2}}\right)^{3}>a_{n}^{3}+3 a_{n}^{3} \cdot \frac{1}{a_{n}^{2}}=a_{n}^{3}+3 \\
\Rightarrow a_{n+1}^{3}-a_{n}^{3}>3 \\
\Rightarrow a_{n+1}^{3}=\sum_{k=1}^{n}\left(a_{k+1}^{3}-a_{k}^{3}\right)+a_{1}^{3}>3 n+a_{1}^{3}>3 n \\
\Rightarrow a_{n+1}>\sqrt[3]{3 n}
\end{array}
$$

Therefore, the sequence $\left\{a_{n}\right\}$ is unbounded.