1000 (-8/9)^(n -1) > 1
(-8/9)^(n -1) is only positive when n is odd
So we can write
1000 (8/9)^(n -1) > 1
(8/9)^(n - 1) > 1/1000 take the log of both sides and by a log property we can write
(n -1) log (8/9) > log (1/1000)
{ divide both sides by log (8/9)....this is negative so reverse the inequality sign }
n - 1 < log (1/1000) / (log (8/9)
n < 1 + log (1/1000) /log (8/9)
n < 59.6
So
n = 59
The number of terms > 1 = [ 59 + 1 ] / 2 = 30