Question 1036781
If r = 0, then |r| <= 1 is certainly true making


S = a/(1-r) = 9/(1-0) = 9/1 = 9


So the smallest possible infinite sum is 9. This is where the first term is 9 and the successive terms after the first term are all 0. 


Ie


S = 9+0+0+0+0+0+.... = 9


we're adding infinitely many 0s to the first term 9


If 0 < r < 1, then S will be larger than 9. For instance, if r = 0.1, then


S = a/(1-r) = 9/(1-0.1) = 9/0.9 = 10


so we can see that the sum is larger if r > 0. 


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So again, to summarize, if the common ratio is r = 0 then the infinite sum is 9. 
This is the smallest infinite sum possible.