daniel_i_l
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I need to calculate
<br /> \sum_{i=1}^n \frac{1}{i(i+1)}<br />
useing the fact that:
<br /> \sum_{i=1}^n F(i) - F(i-1) = F(n) - F(0)<br />
now I chose the function
<br /> F(i) = \frac{1}{i} \frac{1}{(i+1)} ... \frac{1}{(i+r)}<br />
so
<br /> F(i)-F(i-1)=(\frac{1}{i}\frac{1}{(i+1)} ... \frac{1}{(i+r-1)})(\frac{1}{(i+r)}-\frac{1}{(i-1)}) <br />
now I want to use that to calcualte the sum chooseing r as 2, but I'm stuck because the F(0) is undefined, and because of the
<br /> \frac{1}{i+r}-\frac{1}{i-1}) <br />
<br /> \sum_{i=1}^n \frac{1}{i(i+1)}<br />
useing the fact that:
<br /> \sum_{i=1}^n F(i) - F(i-1) = F(n) - F(0)<br />
now I chose the function
<br /> F(i) = \frac{1}{i} \frac{1}{(i+1)} ... \frac{1}{(i+r)}<br />
so
<br /> F(i)-F(i-1)=(\frac{1}{i}\frac{1}{(i+1)} ... \frac{1}{(i+r-1)})(\frac{1}{(i+r)}-\frac{1}{(i-1)}) <br />
now I want to use that to calcualte the sum chooseing r as 2, but I'm stuck because the F(0) is undefined, and because of the
<br /> \frac{1}{i+r}-\frac{1}{i-1}) <br />
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