Let \(\sum_{n = 0}^{123456789} \frac{3n^2 + 9n + 7}{(n^2 + 3n + 2)^3} = \frac{a}{b},\) where \(a\) and \(b\) are relatively prime positive integers. Find \(b-a.\)

Guest Mar 22, 2020

#3**0 **

**This is what Wolfram/Alpha gives as the sum which appears to be an irrational number !!!**

**https://www.wolframalpha.com/input/?i=%E2%88%91%5B+%283n%5E2+%2B+9n+%2B+7%29+%2F+%28n%5E2+%2B+3n+%2B2%29%5E2%2C+n%2C+0%2C+123456789+%5D**

**∑[ (3n^2 + 9n + 7) / (n^2 + 3n +2)^2, n, 0, 123456789 ] =3.289868109.............etc.**

Guest Mar 22, 2020