... 1, 1 2, 1 3, 1 4, . . . , 1 N, . . . ln 1 1 − z=N 1 zNN0, 1, 1 + 1 2, 1 + 1 2+ 1 3, . . . , HN, . . . 1 1 − zln 1 1 − z=N 1 HNzN0, 0, 1, 3 1 2+ 1 3, 4 1 2+ 1 3+ 1 4, ... . 1 (1 − z)M +1 =N≥0N + MNzN 1, 0, 1, 0, . . . , 1, 0, . . . 1 1 − z2=N≥0z2N 1, c, c2, c3, . . . , cN, . . . 1 1 − cz=N≥0cNzN 1, 1, 1 2!, 1 3!, 1 4!, . . . , 1 N!, ... values of the recurrence to get some idea of its rate of growth; try telescoping (iterating) it to get an idea of the asymptotic form of the solution; perhaps look for a summation factor, change of...