MHB [ASK] Find 1/(1)+1/(1+2)+1/(1+2+3)+…+1/(1+2+3+…+2009)

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The discussion focuses on finding the sum of the series 1/(1) + 1/(1+2) + 1/(1+2+3) + ... + 1/(1+2+3+...+2009). Participants note that the denominators represent arithmetic series, which can be simplified using the formula for the sum of the first n integers, S = n(n+1)/2. The series can be rewritten as S = 2(1/(1*2) + 1/(2*3) + 1/(3*4) + ... + 1/(2009*2010). This transformation suggests that there may be a way to simplify the calculation further. The discussion emphasizes the potential for cancellation and shortcuts in evaluating the series.
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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?
 
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Monoxdifly said:
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:

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