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.
Monoxdifly
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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:

$$S=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2009\cdot2010}\right)$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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