Question 486415
Use the fact that


*[tex \LARGE \sum_{i=1}^n \frac{1+2...+i}{1^3 + 2^3 + ... + i^3} = \sum_{i=1}^n \frac{1+2+...+i}{(1+2+...i)^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


*[tex \LARGE \sum_{i=1}^n \frac{2}{i(i+1)} = 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:


*[tex \LARGE 2\sum_{i=1}^n \frac{1}{i} - \frac{1}{i+1} = 2(1 - \frac{1}{n})].