4. Given the geometric sequence $\left\{a_{n}\right\}: a_{1}=5, a_{4}=625$. Then $\sum_{k=1}^{2014} \frac{1}{\log _{5} a_{k} \cdot \log _{5} a_{k+1}}=(\quad)$.
(A) $\frac{2014}{2015}$
(B) $\frac{2013}{2014}$
(C) $\frac{2012}{4028}$
(D) $\frac{2013}{4030}$
4. A.

Let the common ratio of the geometric sequence be $q$. Then, according to the problem,
$$
\begin{array}{l}
625=5 q^{3} \Rightarrow q=5 \\
\Rightarrow \sum_{k=1}^{2014} \frac{1}{\log _{5} a_{k} \cdot \log _{5} a_{k+1}}=\sum_{k=1}^{2014} \frac{1}{k(k+1)}=\frac{2014}{2015} .
\end{array}
$$