Discussing the Convergence of a Series: Get My Opinion!

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SUMMARY

The discussion centers on the convergence of the series defined by the expression \(\frac{n^2 + 3n + 1}{n^2 + 5n + 7}\). Participants clarify that this series does not converge to a geometric series, as it involves a function of \(n\) raised to a power. The limit \(\lim_{n \to \infty}\left( \frac{n^2 + 3n + 1}{n^2 + 5n + 7}\right)^{n^2}\) is critical for determining divergence, particularly when it takes the indeterminate form \([1^\infty]\). The use of logarithms and L'Hopital's Rule is recommended for further analysis.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with limits and indeterminate forms
  • Knowledge of logarithmic functions and their properties
  • Experience with L'Hopital's Rule
NEXT STEPS
  • Study the application of L'Hopital's Rule in limit evaluation
  • Learn about geometric series and their convergence criteria
  • Explore the properties of logarithmic functions in calculus
  • Investigate series convergence tests, such as the Ratio Test and Root Test
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Mathematicians, students studying calculus, and anyone interested in series convergence analysis will benefit from this discussion.

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.
 
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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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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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