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 \(b-a.\)
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.
