5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{n+1}=a_{n}+2 n(n=1,2, \cdots) . \\
\text { Then } \sum_{k=1}^{2007} \frac{1}{a_{k+1}-1}=
\end{array}
$$
5. $\frac{2007}{2008}$.

It is easy to see that $a_{2}=3$.
When $n=1$,
$$
a_{2}-1=1+2-1=2=1 \times(1+1) \text {. }
$$

Assume $a_{k}-1=(k-1) k$.
When $n=k$,
$$
\begin{array}{l}
a_{k+1}-1=a_{k}+2 k-1 \\
=(k-1) k+2 k=k(k+1) .
\end{array}
$$

Thus, $\frac{1}{a_{k+1}-1}=\frac{1}{k(k+1)}$.
Then $\sum_{k=1}^{n} \frac{1}{a_{k+1}-1}=\sum_{k=1}^{n} \frac{1}{k(k+1)}=1-\frac{1}{n+1}$.
Therefore, $\sum_{k=1}^{2007} \frac{1}{a_{k+1}-1}=1-\frac{1}{2007+1}=\frac{2007}{2008}$.