# 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 sum does not seem to simplify unless you accept the use of harmonic numbers. – Start wearing purple Oct 4 '14 at 0:07
• if $m=503-n$, then $$\dfrac{1}{n+505}=-\dfrac{1}{m-1008}$$ So you can still simplify, getting $$\dfrac{1}{3026}\sum\limits_{n=505}^{1008}\dfrac{1}{n}$$ But you can't obtain more.. – Exodd Oct 4 '14 at 0:10
• Oh good to know, thank you :) I thought the same by looking at wolfram result but kept wondering why this was given in telescopic series part of the book – rsadhvika Oct 4 '14 at 0:14
• I was expecting a trick here since the numbers include some suspiciously recent year-numbers... – Omnomnomnom Oct 4 '14 at 0:14

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 $$\frac 1{3026} \ln\left(\frac{1008}{505}\right)$$
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.
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 $$\frac{\log(2)}{3026}=0.000229063840$$
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 $0.0002289$
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 $\approx 0.6926513948612855562$