TITLE: Sum of reciprocals of triangular numbers and calculus
QUESTION [5 upvotes]: I've been searching for interesting calculus homework problems recently and came across the following:
Partition the unit square $[0,1] \times [0,1]$ into regions using the curves $y=x^n$ for $n \in \{0,1,2,\ldots\}$. Then, the area between any two consecutive curves is $$\int_0^1 x^{n-1}-x^n \, dx =\frac{1}{n(n+1)}.$$
Since the area of the unit square is $1$, we get this nice sum: $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1$$
This is where the homework problem ends, but now notice that:
$$\frac{1}{n(n+1)}=\frac{1}{2 {n+1 \choose 2}}$$
Thus: $$ \sum_{n=1}^{\infty}\frac{1}{{n+1 \choose 2}}=2.$$
To put it more gracefully: $$ \sum_{n=2}^{\infty}\frac{1}{{n \choose 2}}=2.$$
Here is what I am wondering: The series given in the last line seems extremely combinatorial. Does anyone know of any combinatorial interpretation of the series or its partial sums? 
I am also wondering if it is possible to derive any other infinite series involving reciprocals of binomial coefficients by using similar techniques.

REPLY [3 votes]: This isn't combinatorial
but algebraic.
However,
generalizations are easier
and encouraged.
$\begin{array}\\
\sum_{n=a}^{b}\dfrac{1}{{n \choose 2}}
&=\sum_{n=a}^{b}\dfrac{1}{n(n-1)/2}\\
&=2\sum_{n=a}^{b}\dfrac{1}{n(n-1)}\\
&=2\sum_{n=a}^{b}\left(\dfrac1{n-1}-\dfrac1{n}\right)\\
&=2\left(\dfrac1{a-1}-\dfrac1{b}\right)\\
\end{array}
$
Your case is
$a=2, b=\infty$.
This can be generalized to
${n \choose m}$
in the denominator.