4. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2, a_{n+1}=\frac{2(n+2)}{n+1} a_{n}\left(n \in \mathbf{N}^{*}\right)$, then $\frac{a_{2014}}{a_{1}+a_{2}+\cdots+a_{2013}}=$
Answer: $\frac{2015}{2013}$.
Solution: Given $a_{n}=\frac{2(n+1)}{n} a_{n-1}=\frac{2(n+1)}{n} \cdot \frac{2 n}{n-1} a_{n-2}=\cdots$
$$
=\frac{2(n+1)}{n} \cdot \frac{2 n}{n-1} \cdots \cdots \frac{2 \cdot 3}{2} a_{1}=2^{n-1}(n+1) .
$$

Let the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $S_{n}$, then

Therefore,
$$
\begin{array}{l}
S_{n}=2+2 \times 3+2^{2} \times 4+\cdots+2^{n-1}(n+1), \\
2 S_{n}=2 \times 2+2^{2} \times 3+2^{3} \times 4+\cdots+2^{n}(n+1),
\end{array}
$$

Subtracting the above two equations, we get
$$
\begin{aligned}
S_{n} & =2^{n}(n+1)-\left(2^{n-1}+2^{n-2}+\cdots+2+2\right) \\
& =2^{n}(n+1)-2^{n}=2^{n} n
\end{aligned}
$$

Thus, $\frac{a_{2014}}{a_{1}+a_{2}+\cdots+a_{2013}}=\frac{2^{2013} \times 2015}{2^{2013} \times 2013}=\frac{2015}{2013}$.