+98 What is the smallest integer that can possibly be the sum of an infinite geometric series whose first term is \(9\)?
+128408
I believe this is correct......
When r = -1/2 the series sums to 9 / [ 1 - (-1/2)] = 9 / (3/2) = 18/ 3 = 6
They guessed at it as follows: divide the first term, 9 in this case, by one of the integers from 1 and up and see if it gives a ratio that converges. So, 9/6 =1.5 - 1 =0.5. Because of the formula for the sum of an infinite series, which is: S = F / [1 - r], they had to make it negative to converge to the smallest integer possible. Hence, 9/[1 - (-1/2)] =9/[1 + 1/2] =6. The same for 4/5 =9/[1 - (-4/5)] =9/[1 + 4/5] =5.
Theoretically, you could choose -8 and you would get: 9/[1 - (-8)] =9/[1 +8] =1, the smallest positive integer.
+128408
Here's the way to figure this.....
Let S = 9 / [ 1 + r]
Since l r l < 1 , then 1 + l r l < 2
Look at the graph of these , here :
https://www.desmos.com/calculator/b5eiddmoz7
Substituting "x" for "r", "S" reaches a positive integer minimum of 5 inside the shaded area when x = .8 = r
Thus
S = 9 / [ 1 + .8 ] = 5
And....re-writing this in a slightly different manner, we have that
S = 9 / [ 1 - r ]
S = 9 / [ 1 - (- .8)] = 9 / [ 1 - (-4/5) ]
So..... r = -4 / 5
