Question

For what r does `3/(n^(2r - 3))-3/n` converge or diverge?

Answer 1

The series sum converges only for `r = 2`

Explanation:

This series can be decomposed as the sum of two series: `sum_n a_n^1 = sum_n 3/(n^{2r-3}` and `sum_n a_n^2 = sum_n -3/n` so that `sum_n a_n = sum_n (a_n^1+a_n^2)`. The series `sum_n a_n^2` is allways divergent and `sum_n a_n^1` converges or diverges depending on `r`. So `sum_n a_n` is convergent only for `r = 2` when `a_n^1+a_n^2=0`