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Help Finding Whether A Sequence Converges

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Q1 Are the following sequences divergent or convergent as n tends to infinity.

    A: [itex]\frac{5n+2}{n-1}[/itex]

    B: [itex]tan^{-1}(n)[/itex]


    2. Relevant equations



    3. The attempt at a solution

    Really not sure how to show this mathematically, or even if what I have done is correct.

    Part A:
    [tex]
    \frac{5n+2}{n-1}=12,\frac{17}{2},\frac{22}{3},\frac{27}{4}...\\[/tex]
    So it looks as tho it converges to 0 as n tends to infinity.

    Part B:
    [tex]
    tan^{-1}(n) = \frac{\pi}{4},1.107,1.25...
    [/tex]
    From messing on the calculator I can see that it tends to pi/2 as n tends to infinity but dont know how to show it mathmatically.

    I would really appreciate any feedback or advice :)
     
  2. jcsd
  3. Oct 14, 2013 #2

    Ray Vickson

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    You have three very similar threads, and in all of them you have shown no effort whatsoever to deal with the problems. Apply the concepts you have learned in class (or should have learned). Is there something about the material you do not understand? If so, tell us what it is.
     
  4. Oct 14, 2013 #3
    There are three threads as they were in one but was told to split them up.

    I am finding the class very difficult, I have given the problems my best effort. I would rather know what I have done wrong so I can see how it should be set out.

    I am not looking for someone to give me the answer. Just some guidance one what I have done wrong and how to set the problem out properly.
     
  5. Oct 14, 2013 #4

    haruspex

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    What do you think ##\frac{5n+2}{n-1}## will tend to for very large n?
     
  6. Oct 14, 2013 #5
    OK I think I have made some progress on my understanding of what the question is actually after.

    Does this make sense?

    A:
    [tex]\frac{5n+2}{n-1}[/tex]
    With this one to numerator will always be bigger, therefore the fraction will keep getting bigger so therefore it diverges to infinity?

    Part B is still confusing me as to why it tends to pi/2 though, is it simply because the tangent function works between 0 and 90° and therefore the limit of the function is 90° which is pi/2?
     
  7. Oct 14, 2013 #6

    Mark44

    Staff: Mentor

    No. Notice that both the numerator and denominator get large as n gets large. The numerator will always be larger, but the whole fraction actually converges to a specific number.
    This is pretty close.
    If we let y = tan-1(n), then n = tan(y).
    What happens to tan(y) as y approaches ##\pi/2## from the left (i.e., for numbers smaller than ##\pi/2##?
     
  8. Oct 14, 2013 #7
    OK thanks, yeah I see that now. From messing on the calculator I can see it converges to 5. I just really dont get how the hell that is shown mathematically though.

    Also for the other one, I can see it gets at least as big as 10381.2, not sure if it gets bigger due to a limit of the number of keys allowed to be entered on my calculator.
     
  9. Oct 14, 2013 #8

    Mark44

    Staff: Mentor

    $$ \frac{5n + 2}{n - 1} = \frac{n(5 + 2/n)}{n(1 - 1/n)} = \frac n n * \frac{5 + 2/n}{1 - 1/n}$$
     
    Last edited: Oct 14, 2013
  10. Oct 14, 2013 #9
    But still from the third bit (the below), how does that show that is will converge to 5? (Sorry, I really am confused by the whole topic at the moment).

    [tex]
    \frac n n * \frac{5 + 2/n}{1 - 1/n}
    [/tex]
     
  11. Oct 14, 2013 #10

    Office_Shredder

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    He meant to write [itex] \frac{n}{n} [/itex] not [itex] \frac{5}{5} [/itex].

    Once you are left with
    [tex] \frac{ 5+2/n}{1-1/n} [/tex]
    as n goes to infinity, what happens to 2/n and 1/n?
     
  12. Oct 14, 2013 #11

    Mark44

    Staff: Mentor

    That should be n/n. I'll fix it.
     
  13. Oct 14, 2013 #12
    Ah, I see. the 2/n and 1/n go to zero and then are left with 5/1 . Thanks.
     
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