Evaluate $\frac{ 1 }{ 1010 \times 2016} + \frac{ 1 }{ 1012 \times 2014} + \frac{ 1 }{ 1014 \times 2012} + \cdots + \frac{ 1 }{ 2016 \times 1010} = ?$ $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = ?
$$

My attempt so far :
$$\sum\limits_{n=0}^{503}\dfrac{1}{(1010+2n)(2016-2n)} =  \dfrac{1}{6052}\sum\limits_{n=0}^{503}\left(\dfrac{1}{n+505} - \dfrac{1}{n-1008}\right)$$
It won't telescope/simplify further. I feel I am in wrong road. Any help ?

The answers already showed that the expression cannot be more simplified. However, it could be quite accurately approximated since $$S_{m,n}=\sum_{i=m}^n \frac{1}{i}=H_n-H_{m-1}$$ Now, consider that both $m$ and $n$ are large numbers; we can use asymptotic expansions and, at the fourth order, $$S_{m,n}\approx \log \left(\frac{n}{m-1}\right)+\frac{1}{2} \left(\frac{1}{n}-\frac{1}{(m-1)}\right)-\frac{1}{12} \left(\frac{1}{n^2}-\frac{1}{(m-1)^2}\right)+\frac{1}{120} \left(\frac{1}{n^3}-\frac{1}{(m-1)^3}\right)+\cdots$$ Using your numbers, the above approximation gives $0.6926513948612855559$ while the summation leads to                       $\approx 0.6926513948612855562$
