Now for the boundary conditions ...

Assuming T(1) = 1, T(1) $\leq$ c1 lg 1 = 0, we cannot choose c large enough.

However, we are not constrained to show for n>=1, but n>=n0.

Extend boundary conditions:

  T(1) = 1              ; Recurrence no longer bottoms out
                        ; at T(1), but at T(2) and T(3)
  T(2) = 2T(1) + 2 = 4
  T(3) = 2T(1) + 3 = 5
  T(2) <= c 2 lg 2 = 2c,    c >= 2
  T(3) <= c 3 lg 3 = 4.75c, c >= 2