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Rational series question

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data

    (n + 1) / (3n - 1)

    2. Relevant equations

    A_n = L

    3. The attempt at a solution

    lim n-> infinity
    (n/n + 1/n) / (3/n - 1/n)

    = (1 + 0) / (3 - 0)

    = 1/3

    Thats the solution, however i have questions..

    1.) If a series is in rational form like this, is it typical to always divide by the largest n in the denominator?

    2.) What is 1/3? Is that the limit of the series? In other words, 1/3 is what it converges to?

    3.) I'm confused as to why the series approaches infinity, so why is 1/3 the limit?

    any help to explain what is going on, would help greatly. Thankyou.
     
  2. jcsd
  3. Apr 9, 2007 #2
    1) yes
    2) 1/3 is the limit of the sequence a_n = (n + 1)/(3n - 1) as n approaches infinity, and yes, we say a_n converges to 1/3
    3) a_n is sequence not a series, and it converges to 1/3 as n->infinity
     
  4. Apr 9, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    To clarify the last part, (3), {an} is not a series (or sum) but a sequence. Further, the sequence does not "approach infinity", it is only the index, n, that "goes to infinity". In fact, {an} decreases steadily from a1= 1 down to 1/3.
     
  5. Apr 10, 2007 #4
    thankyou.

    i have two other problems that i believe both converge if i'm doing them correctly below:

    both n_infinity: 1/(n^2 + 4) and 1/(n^2 + 1)

    so it seems that for both of these problems that the numerator will be 1/n^2
    which is 0? so both converge to 0? just want to make sure they do not diverge.
    thanks

    also......................

    Okay, as for a problem like this: 1/(3n +1)

    where i'm asked to solve it using the intergral test. Do i first want to reduce it like we've done above?

    thanks so far.
     
    Last edited: Apr 10, 2007
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