Discussing the Convergence of a Series: Get My Opinion!

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Homework Help Overview

The discussion revolves around the convergence of a series, specifically examining the behavior of the ratio of polynomials as n approaches infinity. Participants are exploring whether this series can be compared to a geometric series to determine convergence.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish a comparison between two polynomial expressions to argue for convergence. Some participants question the validity of comparing the series to a geometric series, suggesting that the dependence on n complicates the comparison. Others propose using logarithmic transformations and limits to analyze the series further.

Discussion Status

Participants are actively engaging with the original poster's approach, providing alternative perspectives and suggesting methods for analysis. There is a clear exploration of different mathematical techniques, but no consensus has been reached regarding the convergence of the series.

Contextual Notes

Some participants note the presence of an indeterminate form in the limit analysis, which may require specific techniques such as L'Hopital's Rule for resolution. The discussion reflects a mix of established mathematical concepts and ongoing inquiry into the problem's assumptions.

Amaelle
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Homework Statement
studying the convergence of a serie (look at the image)
Relevant Equations
geometric serie, convergence
Good day
I want to study the connvergence of this serie

1612182366542.png

I already have the solution but I want to discuss my approach and get your opinion about it
it s clear that n^2+5n+7>n^2+3n+1 so 0<(n^2+3n+1)/(n^2+5n+7)<1 so we can consider this as a geometric serie that converge?
many thanks in advance
 
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You can't compare to a geometric series; you have a function of n raised to a power which depends on n. That is similar to <br /> \left(1 - \frac 1n\right)^{n} \to e^{-1} &gt; 0. For that reason \sum_{n=1}^\infty \left(1 - \frac1n\right)^n does not converge.

I would write <br /> \frac{n^2 + 3n + 1}{n^2 + 5n +7} = 1 - \frac{2n + 6}{n^2 + 5n +7} and take logs.
 
  • Informative
Likes   Reactions: Amaelle
Amaelle said:
Homework Statement:: studying the convergence of a serie (look at the image)
Relevant Equations:: geometric serie, convergence

Good day
I want to study the connvergence of this serie

View attachment 277247
I already have the solution but I want to discuss my approach and get your opinion about it
it s clear that n^2+5n+7>n^2+3n+1 so 0<(n^2+3n+1)/(n^2+5n+7)<1 so we can consider this as a geometric serie that converge?
many thanks in advance
If you can establish the fact that ##\lim_{n \to \infty}\left( \frac{n^2 + 3n + 1}{n^2 + 5n + 7}\right)^{n^2} \ne 0##, then you can conclude that the series diverges. This limit has the indeterminate form ##[1^\infty]##, so the best way of determining the limit is by the use of logarithms, and getting it to a form in which L'Hopital's Rule can be applied.
 
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Likes   Reactions: Amaelle
pasmith said:
You can't compare to a geometric series; you have a function of n raised to a power which depends on n. That is similar to <br /> \left(1 - \frac 1n\right)^{n} \to e^{-1} &gt; 0. For that reason \sum_{n=1}^\infty \left(1 - \frac1n\right)^n does not converge.

I would write <br /> \frac{n^2 + 3n + 1}{n^2 + 5n +7} = 1 - \frac{2n + 6}{n^2 + 5n +7} and take logs.
thanks you so much!
 
Mark44 said:
If you can establish the fact that ##\lim_{n \to \infty}\left( \frac{n^2 + 3n + 1}{n^2 + 5n + 7}\right)^{n^2} \ne 0##, then you can conclude that the series diverges. This limit has the indeterminate form ##[1^\infty]##, so the best way of determining the limit is by the use of logarithms, and getting it to a form in which L'Hopital's Rule can be applied.
thanks so much!
 

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