What is the smallest integer that can possibly be the sum of an infinite geometric series whose first term is 9?

Sum =F / (1 - r)

1 =9 / (1 - (-8))

1 =9/9 =1 The smallest POSITIVE integer.

Mmmm.......I not certain that the "guest's" answer is correct...if "r"   is  > 1  or < -1, the series diverges and has  no sum

Note:

9 / [ 1 - ( - 1/2)]  = 6

9 / [ 1 - (- 4/5 )]  = 5

So....we are trying to find some "r"   between  -1 and 1 such that

9 / [ 1 - r) ]  = 4       →   r =  -5/4 ......this series will diverge

And

9/ [ 1 - r ]  = 3  →  r = -2.........and this series will diverge, as well

Seemingly, for the positive integer sums of less than 5, "r"  will  be less than -1....and any such series will diverge

So...it appears that the smallest integer that can be the sum of an infinite series whose first term is 9 is produced when r = -4/5.....and that sum is 5

CPhill: I just played with it on W/A and it gave this answer!!

http://www.wolframalpha.com/input/?i=1+%3D+9+%2F+(1+-+r),+solve+for+r

Guest....look at what the progressive sums would be if r = -8

We are evaluating  ∑ 9(-8)^n-1      for n = 1 to infinity

n = 1    sum = 9

n = 2    sum = -68

n = 3  sum =  513

n = 4  sum  = -4095

n = 5  sum = 32,769........and so on

The progressive sums diverge ........!!!!