Sum of reciprocals of triangular numbers and calculus 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.

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.
