SOLUTION: If the common ratio of an infinite series is 3 / 2, it is converging and a sum can be found. True or False Please explain.

Question 1162195: If the common ratio of an infinite series is 3 / 2, it is converging and a sum can be found.
True or False
Please explain.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Answer: False. The infinite series diverges.

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Explanation:

r = 3/2 = 1.5 is larger than 1
when we multiply this common ratio r value by any term, the result is larger than the thing we multiplied 1.5 by

For example, say we multiply 40 with 1.5
We would get 40*1.5 = 60
So the term 40 is followed by 60

Then the next term after 60 is 60*1.5 = 90
and so on

The sequence 40, 60, 90, ... goes on forever without bound. Meaning that the terms just get infinitely large. Adding all infinitely many said terms does not yield a finite sum. The sum of all the terms is infinity. This means the infinite series does not converge to a finite value.

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In contrast, let's say that r is between -1 and 1
In terms of symbols, -1 < r < 1
With this restriction, the geometric sequence would have its terms get smaller and smaller.

Example: r = 1/2 = 0.5 fits the restriction
if we have a first term of 80, then the geometric sequence would be 80, 40, 20, 10, ...
The terms get smaller and smaller tending toward 0. They don't actually get to 0 itself, but they get closer and closer. This is the key to why a sequence like this has its infinite sum converging to a finite value (as opposed to diverging to infinity).

Therefore, if -1 < r < 1, then the infinite geometric series converges. The exact value the sum converges to is S = a/(1-r). With 'a' being the first term and r is the common ratio.

Hopefully this clears everything up.

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