The given conditions lead to: \begin{align} a_n=&\frac{2(n+1)}{n}\cdot a_{n-1}\\ =&\frac{2(n+1)}{n}\cdot\frac{2n}{n-1}\cdot a_{n-2}\\ =&\frac{2(n+1)}{n}\cdot\frac{2n}{n-1}\cdot\frac{2(n-1)}{n-2}\cdot a_{n-3}\\ =&\cdots\\ =&\frac{2(n+1)}{n}\cdot\frac{2n}{n-1}\cdot\frac{2(n-1)}{n-2}\cdots\frac{2\times 3}{2}\cdot a_1\\ =&2^{n-1}(n+1) \end{align} Let $S_n$ be the sum of first $n$ terms of this sequence. Then \begin{align} S_n =& 2 + 2\times 3 + 2^2\times 4 +\cdots +2^{n-1}(n+1)\\ 2S_n =& 2\times 2 + 2^2 \times 3 + 2^3 \times 4 + \cdots + 2^n(n+1) \end{align} Subtracting the $1^{st}$ equation from the $2^{nd}$ one yields: $$S_n = 2^n(n+1)-(2^{n-1}+2^{n-2}+\cdots+2+2)= 2^n(n+1)-2^n = 2^nn$$ Hence, $$\frac{a_{2014}}{a_1+a_2+\cdots+a_{2013}}=\frac{2^{2013}\times (2014+1)}{2^{2013}\times 2013}=\boxed{\frac{2015}{2013}}$$