Series Convergence Test for ∑ 5^n/(4^n +3)

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Homework Help Overview

The discussion revolves around the convergence or divergence of the series ∑ 5^n/(4^n + 3). Participants are exploring the behavior of the series as n approaches infinity, particularly through the application of limits and L'Hôpital's rule.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to evaluate the limit of the series as n approaches infinity and questioning the steps involved in applying L'Hôpital's rule. There is also a discussion on necessary conditions for convergence and alternative methods such as the comparison test.

Discussion Status

The discussion is active, with participants providing insights into the application of L'Hôpital's rule and discussing the implications of the limit not being zero. There are multiple approaches being considered, including the comparison test and geometric series tests.

Contextual Notes

Some participants note that while a limit not equaling zero indicates divergence, it is not a sufficient condition for convergence, prompting further exploration of the series behavior.

bigu01
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Homework Statement


Which of the series, diverge or converge ∑ 5^n/(4^n +3 )


Homework Equations





The Attempt at a Solution

Taking the limit as n→∞ we have (5^n ln 5)/ (4^n ln 4) , my question is here how does it become like this, which part am I missing here?
 
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Oh I see, they have used L'hopital rule since we got infinity over infinity
 
bigu01 said:

Homework Statement


Which of the series, diverge or converge ∑ 5^n/(4^n +3 )


Homework Equations





The Attempt at a Solution

Taking the limit as n→∞ we have (5^n ln 5)/ (4^n ln 4) , my question is here how does it become like this, which part am I missing here?

A necessary condition for \sum a_n to converge is \lim_{n \to \infty} a_n = 0 (it is not a sufficient condition; the series \sum n^{-1} diverges). Here <br /> \lim_{n \to \infty} \frac{5^n}{4^n + 3}<br /> is calculated using L'hopital's rule. Since the limit is not zero the sum does not converge.
 
If ##\Sigma a_n## converges, then ##\lim(a_n) = 0##.

If ##\lim(a_n) ≠ 0##, then ##\Sigma a_n## diverges.

Alternatively, you could apply the comparison test + the geometric test with ##|r| ≥ 1##.
 

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