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A(n+1) = An + 1/(5^n) sequence

  1. Apr 7, 2009 #1
    So, couple sequence questions!

    Converge or Diverge?
    1.
    A(n+1) = An + 1/(5^n) (recursive sequence)

    That should read a sub n plus one equals a sub n plus one over 5 to the n

    I believe it converges, but would like to be sure.


    2. sigma from 1 to infinity of:
    (1+5/n)^n
    That should read the summation from n equals 1 to infinity of one plus 5 over n quantity to the n-th power. I believe this diverges, but would like to be sure.

    Thank you in advance!
     
  2. jcsd
  3. Apr 8, 2009 #2
    Re: Sequences!

    If you have any working I can help you more. What tests do you know about?
     
  4. Apr 8, 2009 #3
    Re: Sequences!

    For a) If [tex] A_{n+1} = A_n + \frac{1}{5^n} [/tex] then [tex] A_{n+1} - A_n = \frac{1}{5^n} [/tex]. Can you prove that this is a Cauchy sequence?

    For b) First prove that [tex] \lim_{n \rightarrow \infty} (1 + c/n)^n = e^c [/tex]. Now, if the summand of the series doesn't have a limit of zero, what can you say about the convergence/divergence of the series?
     
    Last edited: Apr 8, 2009
  5. Apr 8, 2009 #4
    Re: Sequences!

    So doesn't b diverge if you use the n term test?

    And a converges to what, 6? Not sure what a Cauchy sequence is
     
  6. Apr 8, 2009 #5
    Re: Sequences!

    For b, are you familiar with the root test?
     
  7. Apr 8, 2009 #6
    Re: Sequences!

    Nope, but doesn't e^inf not equal 0 meaning it diverges?
     
  8. Apr 8, 2009 #7
    Re: Sequences!

    The root test says that if the limit of the sequence raised to the power 1/n gives a result which is < 1, the series converges.
    If the result is > 1, the series diverges.
    If the result = 1, it can't be determined by this test whether the series converges or diverges.
     
  9. Apr 8, 2009 #8
    Re: Sequences!

    Doesn't n-term work?
     
  10. Apr 9, 2009 #9
    Re: Sequences!

    I don't know if the test I'm talking about is usually called the n-term test but if the summand doesn't have a limit of zero, then the series diverges.

    And a Cauchy sequence is a sequence a_n such that for all epsilon > 0, there is an integer N such that for m,n > N, the following relation holds: |a_m - a_n| < epsilon.

    In your case, you can take m = n+1 and so you must show that [tex] |a_{n+1} - a_n| = \frac{1}{5^n} < \epsilon [/tex]
     
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