Note that $$a_{i+1}+1=2\left(a_{i}+1\right) \text { or } a_{i+1}+1=\frac{a_{i}+a_{i}+2}{a_{i}+2}=\frac{2\left(a_{i}+1\right)}{a_{i}+2}$$ Hence $$\frac{1}{a_{i+1}+1}=\frac{1}{2} \cdot \frac{1}{a_{i}+1} \text { or } \frac{1}{a_{i+1}+1}=\frac{a_{i}+2}{2\left(a_{i}+1\right)}=\frac{1}{2} \cdot \frac{1}{a_{i}+1}+\frac{1}{2}$$ Therefore, $$\frac{1}{a_{k}+1}=\frac{1}{2^{k}} \cdot \frac{1}{a_{0}+1}+\sum_{i=1}^{k} \frac{\varepsilon_{i}}{2^{k-i+1}}$$ where $\varepsilon_{i}=0$ or 1. Multiplying both sides by $2^{k}\left(a_{k}+1\right)$ and putting $a_{k}=2014$, we get $$2^{k}=\frac{2015}{a_{0}+1}+2015 \cdot\left(\sum_{i=1}^{k} \varepsilon_{i} \cdot 2^{i-1}\right)$$ where $\varepsilon_{i}=0$ or 1. Since $\operatorname{gcd}(2,2015)=1$, we have $a_{0}+1=2015$ and $a_{0}=2014$. Therefore, $$2^{k}-1=2015 \cdot\left(\sum_{i=1}^{k} \varepsilon_{i} \cdot 2^{i-1}\right)$$ where $\varepsilon_{i}=0$ or 1. We now need to find the smallest $k$ such that $2015 \mid 2^{k}-1$. Since $2015=5 \cdot 13 \cdot 31$, from the Fermat little theorem we obtain $5\left|2^{4}-1,13\right| 2^{12}-1$ and $31 \mid 2^{30}-1$. We also have $\operatorname{lcm}[4,12,30]=60$, hence $5\left|2^{60}-1,13\right| 2^{60}-1$ and $31 \mid 2^{60}-1$, which gives $2015 \mid 2^{60}-1$. But $5 \nmid 2^{30}-1$ and so $k=60$ is the smallest positive integer such that $2015 \mid 2^{k}-1$. To conclude, the smallest positive integer $k$ such that $a_{k}=2014$ is when $k=60$.