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# Sequences & Series Problem

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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
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#1
+79835
+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
#3
+95
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How did you get $$r=-\frac{1}{2}$$?

benjamingu22  Jun 1, 2017
edited by benjamingu22  Jun 1, 2017
#2
+1

Sum = 9 / [1 -(-4/5)]

Sum = 9 /[1 + 4/5]

Sum = 9 / 1.8

Sum = 5

Guest Jun 1, 2017
#4
+95
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How did you get $$r=-\frac{4}{5}$$?

benjamingu22  Jun 1, 2017
#5
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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
#7
+95
+1

Thank you, Guest.

benjamingu22  Jun 1, 2017
#6
+79835
+1

Thanks, guest....your sum is certainly less than mine....!!!!

CPhill  Jun 1, 2017
#8
+79835
+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
edited by CPhill  Jun 1, 2017
#9
+95
+1

Thank you, CPhill.

benjamingu22  Jun 1, 2017

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