Summation

This problem can be solved neatly by collapsing a telescoping sum.

The $n^{th}$ term of the series is

$\frac{1}{1+2+ \dots +n}=\frac{1}{\frac{1}{2}\left(n\right)\left(n+1\right)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)$

and so the series equals

$2\sum_{n=1}^{n=100}\left(\frac{1}{n}-\frac{1}{n+1}\right)$

Notice that the second term in each summand cancels out the first term in the next summand, so that the sum collapses like a folding telescope to give

$2\left(\frac{1}{1}-\frac{1}{101}\right)=\boxed{1.980}$

Notice that if you extend the series to infinity it converges to 2, which gives the very nice result that 'The sum of the reciptocals of the triangular numbers is equal to two.'!