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 ?
 A: You are almost certainly on the correct road.  We can rewrite this sum as
$$
\frac{1}{6052} \cdot \left(\sum_{n = 0}^{503} \underbrace{\frac 1{n+505}}_{i = n+505} +  \sum_{n = 0}^{503} \underbrace{\frac 1{1008-n}}_{j = 1008-n}\right) =\\
\frac{1}{6052} \cdot \left(\sum_{i = 505}^{1008} \frac 1{i} +  \sum_{j = 505}^{1008} \frac 1{j}\right) = \frac{1}{3026}\sum_{i = 505}^{1008} \frac 1{i}
$$
This doesn't simplify nicely, but it is well approximated by 
$$
\frac 1{3026} \ln\left(\frac{1008}{505}\right)
$$
A: The answer is
$$\frac{H_{1008}-H_{504}}{3026},$$
where $H_n$'s denote harmonic numbers. I don't see, however, how this can be simplified further.
A: First
$$
\frac1{2k(3026-2k)}=\frac1{4\cdot1513}\left(\frac1k+\frac1{1513-k}\right)
$$
So
$$
\begin{align}
\sum_{k=505}^{1008}\frac1{2k(3026-2k)}
&=\frac1{6052}\left(\sum_{k=505}^{1008}\frac1k+\sum_{k=505}^{1008}\frac1{1513-k}\right)\\
&=\frac1{3026}(H_{1008}-H_{504})
\end{align}
$$
Since $\lim\limits_{n\to\infty}(H_{2n}-H_n)=\log(2)$, we have the approximation
$$
\sum_{k=505}^{1008}\frac1{2k(3026-2k)}\approx\frac{\log(2)}{3026}
$$

Evaluating the values above:
$$
\frac1{3026}(H_{1008}-H_{504})=0.000228899998
$$
while
$$
\frac{\log(2)}{3026}=0.000229063840
$$
A: First, note that each term is duplicated  $$\dfrac{ 1 }{ 1010 \times 2016} + \dfrac{ 1 }{ 1012 \times 2014} + \dfrac{ 1 }{ 1014 \times 2012} + \cdots + \dfrac{ 1 }{ 2016 \times 1010} = \\\dfrac{2}{ 1010 \times 2016}+  \dfrac{ 2 }{ 1012 \times 2014} + \dfrac{ 2 }{ 1014 \times 2012} + \cdots + \dfrac{ 2 }{ 1512 \times 1514}=\\\sum_{i=0}^{251}\frac 2{1513^2-(2i+1)^2}$$  Alpha gets a fraction with huge numbers that is about $0.0002289$
A: 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                       $\approx 0.6926513948612855562$
