MHB Converging Geometric Series with Negative Values?

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In a decreasing geometric series, the ratio q must be between 0 and 1 for the series to converge when all terms are positive. If the terms are not all positive, the series can either consist entirely of negative values or alternate in sign. In the case of all negative terms, the negative can be factored into the first term, while alternating series can be expressed with a negative ratio. The convergence condition for a geometric series is that the ratio must satisfy -1 < r < 1. Thus, the behavior of the series changes significantly based on the sign and value of the ratio.
Lancelot1
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Hiya everyone,

Alright ?

I have a simple theoretical question. In a decreasing geometric series, is it true to say that the ratio q has to be 0<q<1, assuming that all members of the series are positive ? What if they weren't all positive ?

Thank you in advance !
 
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A "geometric series" is of the form \sum ar^n= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot= a(1+ r+ r^2+ r^3+ \cdot\cdot\cdot) so, yes, if the series is decreasing and positive then r must be less than 1. If r> 1 then 1&lt; r&lt; r^2&lt; r^3&lt; \cdot\cdot\cdot. If r< 1 then 1&gt; r&gt; r^2&gt; r^3&gt; \cdot\cdot\cdot. Of course, in the first case, r> 1, the series does not converge.

If they are not all positive then either they are all negative and we can take the negative into the "a" term so that \sum at^n= a\sum r^n has a negative number times the same sum of ar^n or they are alternating, \sum a(-r)^n= a\sum (-1)^n r^n. The geometric series \sum ar^n converges if and only if -1&lt; r&lt; 1.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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