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Math Help - Convergence / Divergence

  1. #1
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    Convergence / Divergence

    Determine whether the following infinite series are convergent or divergent

    i) \sum_{n=1}^{\infty }{\frac{n}{n+1}}
    ii) \sum_{n=1}^{\infty }{\frac{n \sqrt{n}}{n^2+1}}
    iii) \sum_{n=2}^{\infty }{\frac{1}{n^2-1}}

    for part 1 and 2 i tired doing ratio test and integral test but they both failed and i got nowhere. anyone want to give me a hand ?

    part 3 seems a lot easier it seems to converge to 0.75 which is easily proved using partial fractions
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobak View Post
    for part 1 and 2 i tired doing ratio test and integral test but they both failed and i got nowhere. anyone want to give me a hand ?

    part 3 seems a lot easier it seems to converge to 0.75 which is easily proved using partial fractions
    for part (i), simply use the test for divergence

    for part (ii), still thinking of the nicest way to do this. don't want to try something like the integral test if it's not necessary

    for part (iii). rewrite \frac 1{n^2 - 1} as \frac A{n + 1} + \frac B{n - 1} (you can use partial fractions, but you don't have to). this should be a telescopic sum. write out some of the terms so you can see which will cancel out and which will stay. take the limit as n \to \infty when done to see the sum
    Last edited by Jhevon; February 18th 2008 at 04:55 PM.
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    part 1 \lim_{n \to{\infty}} \frac{n}{n+1} = 1<br />
so the series is divergent. that makes perfect sense.

    Yeah i got part 3 by the whole method of difference approach i just though i would post it for completeness as it was part of the question
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    Quote Originally Posted by Jhevon View Post
    for part (i), simply use the test for divergence

    for part (ii), still thinking of the nicest way to do this. don't want to try something like the integral test if it's not necessary

    for part (iii). rewrite \frac 1{n^2 - 1} as \frac A{n + 1} + \frac B{n - 1} (you can use partial fractions, but you don't have to). this should be a telescopic sum. write out some of the terms so you can see which will cancel out and which will stay. take the limit as n \to \infty when done to see the sum
    (ii) Note that \frac{n \, \sqrt{n}}{n^2 + 1} \geq \frac{n \, \sqrt{n}}{n^2 + n^2} = \frac{n \, \sqrt{n}}{2n^2} = \frac{1}{2} \, \frac{1}{\sqrt{n}}, for n \geq 1.

    And you know that \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} is divergent.

    From the comparison test, the divergence of \sum_{n=1}^{\infty} \frac{n \, \sqrt{n}}{n^2 + 1} is easily seen.
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    Quote Originally Posted by mr fantastic View Post
    (ii) Note that [tex]

    And you know that \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} is divergent.
    How do we know this ?
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    Quote Originally Posted by bobak View Post
    How do we know this ?
    Because,
    \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}  {\sqrt{n}} \geq \int_1^{n+1} \frac{dx}{\sqrt{x}}  = 2\sqrt{n+1} - 2.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobak View Post
    How do we know this ?
    it is a divergent p-series

    \sum \frac 1{n^p} converges if p>1, diverges if p \le 1

    TPH showed you a more rigorous/specific proof, but this is what it summarizes to in general
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