I saw this problem initially and thought it would be pretty easy from some nice cancellation, but it just didn't happen:
$$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \ldots + \frac{1}{8010}$$
This can be rewritten as:
$$\frac{1}{(1)(2)} + \frac{1}{(2)(3)} + \frac{1}{(3)(4)} + \ldots + \frac{1}{(89)(90)}$$
$$\sum_{n=1}^{89} \frac{1}{n(n+1)}$$
From here, I simply found a pattern by figuring out a few of the first partial sums; they are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}.$ So, the rule for the sum of the first n terms is probably just $\frac{n}{n+1}$, meaning the answer is $\frac{89}{90}$.
However, I want to know how to solve this sum without just calculating the first few partial sums. Is there a formula for this type of series that I just don't know? Or is there a subtle trick I can do with the original sum of fractions, something like in a telescoping series? Thank you.
This is both a generalization and a comment.
Since
$\begin{array}\\ \dfrac1{n(n+1)...(n+m)}-\dfrac1{(n+1)(n+2)...(n+m+1)}\\ =\dfrac1{(n+1)...(n+m)}\left(\dfrac1{n}-\dfrac1{(n+m+1)}\right)\\ =\dfrac1{(n+1)...(n+m)}\left(\dfrac{m+1}{n(n+m+1)}\right)\\ =\dfrac{m+1}{n(n+1)...(n+m)(n+m+1)}\\ \end{array} $
replacing $m+1$ by $m$
$\dfrac{m}{n(n+1)...(n+m)(n+m)} =\dfrac1{n(n+1)...(n+m-1)}-\dfrac1{(n+1)(n+2)...(n+m)} $
so
$\dfrac1{n(n+1)...(n+m)(n+m)} =\dfrac1{m}\left(\dfrac1{n(n+1)...(n+m-1)}-\dfrac1{(n+1)(n+2)...(n+m)}\right) $
Therefore
$\begin{array}\\ \sum_{n=a}^b \dfrac1{n(n+1)...(n+m)(n+m)}\\ =\sum_{n=a}^b\dfrac1{m}\left(\dfrac1{n(n+1)...(n+m-1)}-\dfrac1{(n+1)(n+2)...(n+m)}\right)\\ =\dfrac1{m}\sum_{n=a}^b\left(\dfrac1{n(n+1)...(n+m-1)}-\dfrac1{(n+1)(n+2)...(n+m)}\right)\\ =\dfrac1{m}\left(\sum_{n=a}^b\dfrac1{n(n+1)...(n+m-1)}-\sum_{n=a}^b\dfrac1{(n+1)(n+2)...(n+m)}\right)\\ =\dfrac1{m}\left(\sum_{n=a}^b\dfrac1{n(n+1)...(n+m-1)}-\sum_{n=a+1}^{b+1}\dfrac1{n(n+1)(n+2)...(n+m-1)}\right)\\ =\dfrac1{m}\left(\dfrac1{a(a+1)...(a+m-1)}-\dfrac1{(b+1)(b+2)...(b+m)}\right)\\ \end{array} $