3. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{2}, a_{n+1}=\frac{a_{n}}{2(n+1) a_{n}+1}$, then $\sum_{k=1}^{2023} a_{k}=$ $\qquad$
$$
\begin{array}{l}
a_{n+1}=\frac{a_{n}}{2(n+1) a_{n}+1} \Rightarrow \frac{1}{a_{n+1}}=\frac{1}{a_{n}}+2(n+1), \text{ then we have } \frac{1}{a_{n}}=\frac{1}{a_{1}}+\sum_{i=1}^{n-1}\left(\frac{1}{a_{i+1}}-\frac{1}{a_{i}}\right) \\
=2+2 \sum_{i=1}^{n-1}(i+1)=2+(n-1)(n+2)=n(n+1) \Rightarrow a_{n}=\frac{1}{n(n+1)} .
\end{array}
$$

Therefore, $\sum_{k=1}^{2023} a_{k}=\sum_{k=1}^{2023} \frac{1}{k(k+1)}=\sum_{k=1}^{2023}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{2024}=\frac{2023}{2024}$.