You can
put this solution on YOUR website!Use the fact that
^2} = \sum_{i=1}^n \frac{1}{1+2+...+i})
Since the sum of the first i positive integers is i(i+1)/2, we reciprocate so our sum is equal to
} = 2\sum_{i=1}^n \frac{1}{i(i+1)})
I learned a nice Mathcounts trick that you can decompose that fraction to 1/i - 1/(i+1) (if you don't know it, just decompose it normally using partial fractions). This becomes a "telescoping" sum where all the terms collapse except the first and last:
)
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