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

benjamingu22
May 31, 2017

#1**+1 **

I believe this is correct......

When r = -1/2 the series sums to 9 / [ 1 - (-1/2)] = 9 / (3/2) = 18/ 3 = 6

CPhill
May 31, 2017

#5**0 **

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.

Guest Jun 1, 2017

edited by
Guest
Jun 1, 2017

#8**+1 **

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

CPhill
Jun 1, 2017