5. In the sequence $\left\{a_{n}\right\}$,
$$
\begin{array}{l}
a_{1}=1 \\
a_{n}=a_{n-1}+\frac{1}{n(n-1)}\left(n \geqslant 2, n \in \mathbf{Z}_{+}\right) .
\end{array}
$$

Then $a_{4}=$ ( ).
(A) $\frac{7}{4}$
(B) $-\frac{7}{4}$
(C) $\frac{4}{7}$
(D) $-\frac{4}{7}$
5. A.

Since $a_{n}-a_{n-1}=\frac{1}{n-1}-\frac{1}{n}$, therefore,
$$
\begin{array}{l}
a_{4}-a_{1}=\sum_{n=2}^{4}\left(\frac{1}{n-1}-\frac{1}{n}\right)=1-\frac{1}{4} \\
\Rightarrow a_{4}=a_{1}+\frac{3}{4}=\frac{7}{4} .
\end{array}
$$