Sum of reciprocals of triangular numbers up to 1+2+…+2009

[Monoxdifly]

Does anyone know how to add these fractions?
[math]\frac11+\frac1{1+2}+\frac1{1+2+3}+…+\frac1{1+2+3+…+2009}[/math]
Like my previous question, I believe this one also has something which can be canceled out, and the denominators contain arithmetic series. Can this series be used to make some sorts of shortcuts?

 

[MarkFL]

Does anyone know how to add these fractions?
[math]\frac11+\frac1{1+2}+\frac1{1+2+3}+…+\frac1{1+2+3+…+2009}[/math]
Like my previous question, I believe this one also has something which can be canceled out, and the denominators contain arithmetic series. Can this series be used to make some sorts of shortcuts?

We know:

$$\sum_{k=1}^n(k)=\frac{n(n+1)}{2}$$

And so the given series becomes:

$$S=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2009\cdot2010}\right)$$