1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Convergent/Divergent Series

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    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)

    2. Relevant 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

    3. The attempt at a solution

    I tried saying that bn was (4/6)^n but i dont know how to compare that to the original series
  2. jcsd
  3. Mar 30, 2009 #2


    User Avatar
    Science Advisor

    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].
  4. Mar 30, 2009 #3
    so then because 2(4/6)^n is convergent, the original series is convergent, correct?
  5. Mar 30, 2009 #4
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook