Summing up the Infinite: 3(1/11)^n

  • Thread starter Thread starter Lance WIlliam
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Homework Help Overview

The discussion revolves around the summation of an infinite geometric series, specifically the series represented by the expression 3(1/11)^n as n approaches infinity. Participants are exploring the convergence and calculation of the sum of this series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the structure of the series and the application of the geometric series formula. There is uncertainty about the correct approach to finding the actual sum, with some participants questioning their understanding of the geometric series properties.

Discussion Status

The conversation is active, with participants offering insights into the geometric series formula and its application. While one participant expresses confusion about calculating the sum, another provides a formula, suggesting a productive direction in the discussion.

Contextual Notes

There is a mention of the geometric series test for convergence, and participants are navigating the implications of starting the summation at n=0. The original poster's initial assumption about the sum is questioned, indicating a need for clarification on the series' behavior.

Lance WIlliam
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\sum as n=0(there's a infinity above that sigma) , 3(1/11)^n

I thought it would just be 3/11 and converge due to the geometric test but its not...to find the sum would I just start at 0 and put numbers in for "n"...
 
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You mean \sum^{\infty}_{n=0} 3\left(\frac{1}{11^n}\right). You can factor the 3 outside of the sigma. And the resulting would be a geometric series, no?
 
yes but I don't see how I would go about finding a actual sum.
 
What do you know about the sum of a geometric series?
 
OH! S=a/1-r
 
its 33/10 thankyou
 

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