Is this Series Convergent or Divergent Using Comparison Tests?

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of a specific series using comparison tests. The series in question is \(\sum_{n=1}^{\infty} \frac{n + 4^n}{n + 6^n}\), which falls under the subject area of series convergence in calculus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to identify a suitable comparison series and questions how to effectively compare it to the original series. Other participants discuss the relationship between the terms of the series and suggest inequalities to establish a comparison.

Discussion Status

Participants are exploring various comparisons and have identified a potential comparison series. There is some agreement on the reasoning presented, but the discussion does not reach a definitive conclusion regarding the convergence of the original series.

Contextual Notes

Participants are working within the constraints of using comparison tests and are examining the behavior of the series as \(n\) approaches infinity. There is an implicit assumption about the convergence of the comparison series being used.

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Homework Statement


I have to determine whether the given series is convergent or divergent using the comparison tests:
[tex]\sum[/tex] from n = 1 to infinity of (n + 4n / (n + 6n)


Homework Equations


If bn is convergent and an [tex]\leq[/tex] bn then an is also convergent

liimit of an/bn as n goes to infinity = c, if c > 0, then both are either convergent or divergent

The Attempt at a Solution



I tried saying that bn was (4/6)^n but i don't know how to compare that to the original series
 
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Certainly [itex](n+4^n)/(n+ 6^n)< (n+ 4^n)/6^n[/itex] because the left side has a larger denominator. It is also true that n< 4^n for any positive integer n. That means that [itex]n+ 4^n< 4^n+ 4^n< 2(4^n)[/itex] and so [itex](n+4^n)/(n+6^n)< (n+4^n)/6^n< 2(4^n)/6^n)[/itex].
 
so then because 2(4/6)^n is convergent, the original series is convergent, correct?
 
Correct.
 

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