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

  • Thread starter Sonic7145
  • Start date
  • #1
58
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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!
 

Answers and Replies

  • #2
283
3


If you have any working I can help you more. What tests do you know about?
 
  • #3
726
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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:
  • #4
58
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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
 
  • #5
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For b, are you familiar with the root test?
 
  • #6
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Nope, but doesn't e^inf not equal 0 meaning it diverges?
 
  • #7
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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.
 
  • #8
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Doesn't n-term work?
 
  • #9
726
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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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