Explaining Convergence of (3^n)/(2^n + 4^n) w/o Limit Comp. Test

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Homework Help Overview

The discussion revolves around the convergence of the series (3^n)/(2^n + 4^n) and the application of the comparison test, specifically avoiding the limit comparison test as per the original poster's request.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore comparisons to geometric series, particularly 1/2^n, and discuss the validity of certain inequalities and manipulations involving the terms of the series.

Discussion Status

There is an ongoing exploration of the appropriate comparisons and manipulations to establish convergence. Some participants express uncertainty about the legality of specific steps taken in their reasoning, indicating a productive dialogue about the problem's setup.

Contextual Notes

Participants are constrained by the requirement not to use the limit comparison test and are questioning the validity of their algebraic manipulations in the context of the comparison test.

Erind
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Homework Statement



Use the comparison test to explain why the series (3^n)/(2^n + 4^n) is convergent. He said specifically not to use the limit comparison test on this one.

Homework Equations



1/2^n

The Attempt at a Solution



I know I should be comparing it to 1/2^n because it is a geometric series and thereby convergent, but I can't get rid of the 3^n on the top in order to get there.
 
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Ok, I just came up with comparing (3^n)/(2^n + 4^n) < (3^n)/(2^n + 4^n)(3^n) and then canceling the (3^n) and then 1/(2^n + 4^n) < 1/(2^n) and then it's done, but I'm not sure if that first inequality is a legal move.
 
No, I'm afraid that it isn't a legal move :frown:

Maybe you should delete some of the terms in the denumerator...
 
Try starting with 2n + 4n > 4n, then rewrite the inequality until you get 3n/(2n + 4n) for the left side of the inequality.
 

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