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Math Help - converge or diverge

  1. #1
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    converge or diverge

    does this series converge or diverge?

    n = 2 E infinity

    2/(n^2 -1)

    ..i've attempted to solve this, and i know the answer is 3/2

    just not sure why i could not divide the numerator and denominator by the largest exponent in the denominator (n^2). is that because its not in the form of infininty/infinity?

    also, tried using summation notation, it did not work out.

    i know it can't be the harmonic series because that would diverge.

    need some help please.
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  2. #2
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    Quote Originally Posted by rcmango View Post
    does this series converge or diverge?

    n = 2 E infinity

    2/(n^2 -1)
    .
    \left| \frac{2}{n^2 - 1} \right| \leq \frac{2}{n^2 - \frac{1}{2}n^2} = \frac{4}{n^2}
    Now use comparasion test.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rcmango View Post
    does this series converge or diverge?

    n = 2 E infinity

    2/(n^2 -1)

    ..i've attempted to solve this, and i know the answer is 3/2

    just not sure why i could not divide the numerator and denominator by the largest exponent in the denominator (n^2). is that because its not in the form of infininty/infinity?

    also, tried using summation notation, it did not work out.

    i know it can't be the harmonic series because that would diverge.

    need some help please.
    to find the sum, you can realize that you have a telescoping sum. note that \frac 2{n^2 - 1} = \frac 1{n - 1} - \frac 1{n + 1}

    write out some of the terms of the series (be sure to include the last few terms as well, that is, say, the (n - 2)th term, the (n - 1)th term, the nth term). try to see a pattern of what cancels out and come up with an expression for what's left. then let n \to \infty and you should get your result
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  4. #4
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    Okay, thanks for the help, i can't used the comparison test because i haven't learned it yet.

    however i can use the telescoping series test, is that the same as the collapsing series?

    also.. i've attached an image of what work i've done, I apoligize it is very messy, sorry was in a hurry.

    btw, i see how the only terms of 1/2 and 1 are left only. but what happened to the (1/n-1) - (1/n+1) do those cancel out completely?

    we used partial sums here to split the fraction up then?

    thanks alot.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rcmango View Post
    Okay, thanks for the help, i can't used the comparison test because i haven't learned it yet.

    however i can use the telescoping series test, is that the same as the collapsing series?
    i don't know. don't think i've heard the term "collapsing series" before.

    also.. i've attached an image of what work i've done, I apoligize it is very messy, sorry was in a hurry.
    where's the image?

    btw, i see how the only terms of 1/2 and 1 are left only. but what happened to the (1/n-1) - (1/n+1) do those cancel out completely?
    this is why i told you to write out some terms. you'd realize that the (n - 2)th term cancels the \frac 1{n - 1}, but the \frac 1{n + 1} stays. you'd have to take the limit as n \to \infty to get the answer

    we used partial sums here to split the fraction up then?
    i used partial fractions decomposition to get the two fractions
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