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anonymous

  • 5 years ago

Can the ratio test imply convergence to zero for a sequence? for example show n! / n^n converges,

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  1. anonymous
    • 5 years ago
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    ahh ratio test is useless here as well

  2. anonymous
    • 5 years ago
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    double whammy

  3. anonymous
    • 5 years ago
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    you get (n+1)! / (n+1)^ n+1 * n^n / n! = n^n / (n+1)^n = (n / n+1)^n

  4. anonymous
    • 5 years ago
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    the limit produces 1^n

  5. anonymous
    • 5 years ago
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    1 ^ infinity, so indeterminate?

  6. anonymous
    • 5 years ago
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    ahh, there is still hope

  7. anonymous
    • 5 years ago
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    You can try the definition of convergence. We want to show that n!/n^n converges to 0. So we’d like to show that for any epsilon > 0, we can find some positive integer M such that for all positive integers N > M, we’d have that N!/N^N is less than epsilon. This would show that n!/n^n converges to 0. Proof: Let epsilon be some arbitrarily small positive number less than 1/2. We know that if we wanted to, we could find some positive integer J such that 0 < 1/J < epsilon, since the sequence {1/n} converges to 0. We wanted to show that we can find some positive integer M such that for all positive integers N > M, we’d have that N!/N^N is less than epsilon. Instead, we will show that we can find some positive integer M such that for all positive integers N > M, we’d have that N!/N^N is less than 1/J. This would imply that N!/N^N < epsilon. My claim is that you can let M be J, because J!/J^J is less than 1/J for all J > 2 (which holds since epsilon is less than 1/2). We can show that J!/J^J < 1/J by showing that J! < J^(J-1), which we can do by induction (omitted). Therefore, for any epsilon > 0, we can find some positive integer M such that for all positive integers N > M, we’d have that N!/N^N is less than epsilon. So n!/n^n converges to 0.

  8. anonymous
    • 5 years ago
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    wow

  9. anonymous
    • 5 years ago
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    what about this idea n!/n^n=1*2*..*(n-1)*n/(n*n*...*n)= =(1/n)*{(2/n)*...*(1-2/n)}*(1-1/n)*1 Each term in braces is less than (1-1/n). There are (n-2) terms there, so n!/n^n<(1/n)*(1-1/n)^(n-2)*1

  10. anonymous
    • 5 years ago
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    now using sandwhich theorem...

  11. anonymous
    • 5 years ago
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    we know that lim r^n for | r | < 1 is 0 , right?

  12. anonymous
    • 5 years ago
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    Yeah I think that n!/n^n<(1/n)*(1-1/n)^(n-2)*1 also works. You'd be able to rewrite the right side as [n/(n-1)^2] * (1 – 1/n)^n, which converges to 0.

  13. anonymous
    • 5 years ago
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    right

  14. anonymous
    • 5 years ago
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    whats an example of a non archimedian ordered field?

  15. anonymous
    • 5 years ago
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    I don't know.

  16. anonymous
    • 5 years ago
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    youre a math genius !!! . to get that epsilon proof thanks

  17. anonymous
    • 5 years ago
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    i am humbled by your solution

  18. anonymous
    • 5 years ago
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    but wait, there is more i can do. but just wanted to ask, the ratio test doesnt help us here, but if the ratio test converges to zero, does that mean that the sequence must converge to zero ( i know the series will converge)

  19. anonymous
    • 5 years ago
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    This is very cool. watch this

  20. anonymous
    • 5 years ago
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    lim n→+∞ [(1 - 1/n)⁽⁻ⁿ⁾]⁻¹ = lim n→+∞ {[(n - 1) /n ]⁽⁻ⁿ⁾}⁻¹ = lim n→+∞ {{1/ [(n - 1) / n ]} ⁿ}⁻¹ = lim n→+∞ {[ n /(n - 1)]ⁿ}⁻¹ = you can easily verify that n / (n - 1) can be written as 1+1/(n - 1), so: lim n→+∞ {[ n /(n - 1)]ⁿ}⁻¹ = lim n→+∞ {[ 1+1/(n -1)]ⁿ}⁻¹ = lim n→+∞ {[1+1/(n -1)]∙[1+1/(n -1)]⁽ⁿ⁻¹⁾}⁻¹ = notice that if n→+∞ then (n -1) →+∞ too, therefore lim n→+∞ [1+1/(n -1)]⁽ⁿ⁻¹⁾→ e and, finally lim n→+∞ [(1+0)∙e]⁻¹ = 1/e

  21. anonymous
    • 5 years ago
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    so we can use this as a lemma in our sequence n! / n^n

  22. anonymous
    • 5 years ago
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    lim n→+∞ (1 - 1/n)ⁿ = 1/e

  23. anonymous
    • 5 years ago
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    n!/n^n<(1/n)*(1-1/n)^(n-2)= [n/(n-1)^2] * (1 – 1/n)^n = 0 * 1/e

  24. anonymous
    • 5 years ago
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    yep, both pretty slick solutions

  25. anonymous
    • 5 years ago
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    actually we need that lemma, right , i think. or .. since it is not lim r^n for |r|< 1 , not quite the same thing

  26. anonymous
    • 5 years ago
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    and then here i used sandwich, 0< n!/n^n<(1/n)*(1-1/n)^(n-2)= [n/(n-1)^2] * (1 – 1/n)^n = 0 * 1/e hmmm, can you sandwich sequences, i dont know

  27. anonymous
    • 5 years ago
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    0< n!/n^n < 0 in the limit as n goes to infinity i bet there is a sandwich theorem for sequences

  28. anonymous
    • 5 years ago
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    interesting we have lim n→+∞ (1 - 1/n)ⁿ = 1/e and lim n→+∞ (1 + 1/n)ⁿ = e

  29. anonymous
    • 5 years ago
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    watch out, this website will freeze your work i type it in notepad first

  30. anonymous
    • 5 years ago
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    you have to be careful to keep the limits. otherwise, you get nonsense results like 0 < 0. we showed that n!/n^n < [n/(n-1)^2] * (1 – 1/n)^n So taking the limit as n goes to infinity of both sides gives lim n!/n^n <= lim [n/(n-1)^2] * (1 – 1/n)^n But the right side equal lim [n/(n-1)^2] * lim (1 – 1/n)^n = 0 * e = 0. Since values of n are all positive, it follows that lim n!/n^n = 0.

  31. anonymous
    • 5 years ago
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    0 * e^-1

  32. anonymous
    • 5 years ago
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    ....you do realize it's getting complicated here for someone that is probably not at the level of using epsilon/N definition in their HW. Just use the ratio test and you get (n/n+1)^n like you said. So then... \[(n/n+1)^n = (1/(1+1/n))^n\] We see that this approaches 1/e which is less than 1. So this series converges

  33. anonymous
    • 5 years ago
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    nope that wont work

  34. anonymous
    • 5 years ago
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    Well, the author's screen name was cantorset, so I assumed he's taking upper division math classes in college. The proof is pretty standard in those classes. The problem is asking us to find what a sequence converges to, not what the sum converges to. Ratio tests for sequences can show that a sequence is decreasing, but not necessarily what it's decreasing to.

  35. anonymous
    • 5 years ago
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    the series converges , not necessarily the sequence. Im trying to think of a counterexample...

  36. anonymous
    • 5 years ago
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    right, we wanted to show that the sequence converges to a particular value , here zero. so the ratio test is of no luck here

  37. anonymous
    • 5 years ago
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    No, the ratio test is simply to determine if the series is convergent. The original question is asking for if the series do infact converges. Not the series.

  38. anonymous
    • 5 years ago
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    nope, the original question was for the sequence n! / n^n, show that it converges to zero

  39. anonymous
    • 5 years ago
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    ....the author doesn't know how to put a sigma sign....It's not that complicated of a question. Trust me

  40. anonymous
    • 5 years ago
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    right, the ratio test shows that a positive sequence is decreasing. because of the condition L < 1

  41. anonymous
    • 5 years ago
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    It's one of the fundamental mistakes people make to be confused between sequence and series. Cause often they treat it like the same thing.

  42. anonymous
    • 5 years ago
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    Given that the sequence is positive, decreasing, and bounded below by zero. the series must converge to some value . but by the divergence test since the seris converges then the lim a_n must go to zero. if lim a_n does not to go zero then the series diverges. so the contrapositive i think solves this problem

  43. anonymous
    • 5 years ago
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    well we need to prove the ratio test here. but i think we can use the divergence test (the contrapositive) because the series n! / n^n converges by ratio test, then the limit of the sequence must go to zero . remember divergence test?

  44. anonymous
    • 5 years ago
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    Yes that works. Since the series converges, then limit must be 0

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