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Homework Help: Determine whether the series converges or diverges

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series converges or diverges.

    [tex]\sum[/tex](1+5^n) / (1+6^n)

    2. Relevant equations

    limit comparison test, or just the comparison test.

    3. The attempt at a solution

    [attempt #1]

    I have a couple of ways of trying to proves this.

    1) (1+5^n) / (1+6^n) < (1 + 5^n) / 6^n =

    (1/6)^n + (5/6)^n , which converges to 1/5 + 5 = 26/5

    so since the series is bounded by 0 <= (1+5^n)/(1+6^n) <=26/5
    it converges? This might not necessarily be true because a function might fluctuate between
    those points until infinity, but I think that only cosine have this property. Thus this series converges.

    [attempt # 2]

    (1+5^n) / (1+6^n) =

    1 / (1+6^n) + (5/6)^n

    I know that (5/6)^n converges to 5, so

    = 1/(1+6^n) + 5

    now I apply the comparison test to 1/(1+6^n)

    1/(1+6^n) < 1/(6^n), and this converges to 0.2

    so ,

    [0<= 1/(1+6^n) <= 0.2 ] + 5

    so it converges since 1/(1+6^n) is bounded.

    Either 1 right?
  2. jcsd
  3. Mar 22, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Re: Converge/diverge

    Yes, use a comparison test. The first attempt is better. But if you are going to give values for the sum you should probably give a value for the lower limit on n. Is it 0 or 1 or 2 or what?
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