5. Let $f(x)$ be a polynomial of degree 2014,

such that $f(k)=\frac{1}{k}(k=1,2, \cdots, 2015)$. Then $f(2016)=$ $\qquad$ - .
Construct the function $g(x)=x f(x)-1$. According to the problem, $g(1)=g(2)=\cdots=g(2015)=0$. Clearly, $g(x)$ is a polynomial of degree 2015, so $g(x)=a(x-1)(x-2) \cdots(x-2015)$. Therefore, $f(x)=\frac{a(x-1)(x-2) \cdots(x-2015)+1}{x}$. Since $f(x)$ is a polynomial of degree 2014, the constant term of $a(x-1)(x-2) \cdots(x-2015)+1$ must be 0, i.e., $(-1)^{2015} 2015!\cdot a+1=0 \Rightarrow a=\frac{1}{2015!}$.
Thus, $f(x)=\frac{\frac{(x-1)(x-2) \cdots(x-2015)}{2015!}+1}{x} \Rightarrow f(2016)=\frac{2}{2016}=\frac{1}{1008}$.