3. The function $f(x)$ satisfies $f(1)=1$, and
$$
f(n)=f(n-1)+\frac{1}{n(n-1)}\left(n \geqslant 2, n \in \mathbf{Z}_{+}\right) \text {. }
$$

Then $f(2019)=$
3. $\frac{4037}{2019}$.
$$
\begin{array}{l}
\text { Given } f(n)-f(n-1) \\
=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n},
\end{array}
$$

we know
$f(2)-f(1)=1-\frac{1}{2}$,
$f(3)-f(2)=\frac{1}{2}-\frac{1}{3}$,
$\qquad$
$f(2019)-f(2018)=\frac{1}{2018}-\frac{1}{2019}$.
Adding the above equations on both sides, we get
$$
\begin{array}{l}
f(2019)-f(1)=1-\frac{1}{2019} \\
\Rightarrow f(2019)=2-\frac{1}{2019}=\frac{4037}{2019} .
\end{array}
$$