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Math Help - convergent or divergent series

  1. #1
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    convergent or divergent series

    are the following sums,with general tern an, convergent or divergent.

    an=(1+n)/n^2

    my book says convergent but i think divergent since

    an=1/n^2 + 1/n >1/n

    2)

    an=(1+n)/(2+n^2)

    book says convergent but i think divergent

    this time i think an>(1+n)/(n+n^2)=1/n

    3) an=1/[n ln(n)] convergent

    compare with un=1/n^2 so an/un=n/ln(n) which tends to 1

    4) an=ln(n)/n^2..
    i think convergent but not sure how to show this.


    5) an=ln (n)/(2n^3-1)
    as for 4. think convergent but not doing correct comparison.


    5) an=(2n-1)/n(n+1)(n+2)
    this converges, i put into partial fractions and got an expression for the sum. Is there a nice comparison to show it quicker?
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  2. #2
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    Quote Originally Posted by jiboom View Post
    are the following sums,with general tern an, convergent or divergent.

    an=(1+n)/n^2

    my book says convergent but i think divergent since

    an=1/n^2 + 1/n >1/n
    \sum\frac{1+n}{n^2}=\sum\frac{1}{n^2}+\sum\frac{1}  {n}=\infty.

    So either your book is wrong or else you are misreading it.

    2)

    an=(1+n)/(2+n^2)

    book says convergent but i think divergent

    this time i think an>(1+n)/(n+n^2)=1/n
    \frac{n^2}{2+n^2}\geq \frac{1}{3} for positive n, and so

    \frac{1+n}{2+n^2}\geq\frac{n}{2+n^2}\geq\frac{1}{3  n}.

    Therefore \sum\frac{1+n}{2+n^2}=\infty.

    Again, either your book is wrong or else you're misreading it.

    3) an=1/[n ln(n)] convergent

    compare with un=1/n^2 so an/un=n/ln(n) which tends to 1
    It is not true that \frac{n}{\ln n}\to 1.

    Instead, use the integral test. Letting u=\ln x, we have

    \int_C^{\infty}\frac{1}{\ln x}(1/x)dx=\int_{C'}^{\infty}u^{-1}du=\left[\ln u\right]_{C'}^{\infty}=\infty.

    So \sum\frac{1}{n\ln n} diverges.

    4) an=ln(n)/n^2..
    i think convergent but not sure how to show this.
    Notice that \ln n\leq \sqrt{n} for positive n. So

    \frac{\ln n}{\sqrt{n}}\leq1, and

    \frac{\ln n}{n^2}\leq\frac{1}{n^{3/2}}.

    Therefore \sum\frac{\ln n}{n^2}\leq\sum\frac{1}{n^{3/2}}<\infty.

    5) an=ln (n)/(2n^3-1)
    as for 4. think convergent but not doing correct comparison.
    This does indeed follow from (4).

    5) an=(2n-1)/n(n+1)(n+2)
    this converges, i put into partial fractions and got an expression for the sum. Is there a nice comparison to show it quicker?
    Use inequalities instead of trying to multiply out. For example,

    \frac{2n-1}{n(n+1)(n+2)}\leq\frac{2n}{n(n+1)(n+2)}\leq\frac  {2n}{n^3}=2\frac{1}{n^2}.

    And of course the comparison test will work quite easily here.
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  3. #3
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    thanks for the confirmation on 1 and 2. The answer in book says convergent, so i was worried i was getting divergent. Although in another question it gives the answer as divergent but then later on in the exercise asks to show the series is convergent

    Sorry for the bad error, i thought ln(n)/n was not tending to 1 but graphed it and it stayed at 1. Thats an unistall to do later

    Im afraid i dont have integral test at my disposal, only comparison and ratio,so is there a way to do that one without integral test?

    edit: got mixed up with an=ln(n)/n !!!

    Finally, how do i show ln x<rt(x)? My book only gets me to show ln(x)<x in the next question, which does not help for
    Many thanks.
    Last edited by jiboom; February 3rd 2010 at 05:06 AM.
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  4. #4
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    i am still unable to find a decent comparison to do an=1/(nln(n)). i either get a 1 with ratio test or tending to 0 with a comparison.

    Another question which i dispute the book answer. They have divergent.

    here we consider convergence of

    \sum ne^{(-n^2)}<br /> <br />

    im saying:



    \frac{n}{e^{(n^2)}} \leq \frac{n}{e^{(n)}}

    using ratio test on



    \sum ne^{(-n)}<br /> <br />



    gives



     <br /> <br />
\frac{a_{n+1}}{a_{n}} = \frac{(n+1)e^{n}}{ne^{(n+1)}}



     <br /> <br />
=\frac{1}{e^{1}}[1+\frac{1}{n}]<br /> <br />



    which tends to e^{(-1)}

    so

    \sum ne^{(-n)}<br /> <br />

    converges and hence so does

    \sum ne^{(-n^2)}<br /> <br />



    finally still not sure how to get ln(x)<rt(x)
    Last edited by jiboom; February 3rd 2010 at 02:39 PM.
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  5. #5
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    Quote Originally Posted by jiboom View Post
    i am still unable to find a decent comparison to do an=1/(nln(n)). i either get a 1 with ratio test or tending to 0 with a comparison.

    ANother question which i dispute the book answer. They have divergent.

    here an=n(exp(-n^2)

    im saying:

    n(exp(-n^2))<n(exp(-n)
    and using ratio test

    an+1/an
    =[(n+1)exp(n)]/[nexp(n+1)]
    =1/exp(1)[1+1/n]
    which tends to exp(-1)
    so sum n(exp(-n) converges and hence so does sum (an)

    finally still not sure how to get ln(x)<rt(x)
    ok,got the ln(x)<rt(x), just considered y=rt(x)-ln(x) and found no turning points etc.

    Im still unable to find a comparison/ratio test for \sum \frac{1}{nln(n)}
    Been on google looking but nothing. can it be done by these methods?

    is my answer of convergent correct for
    \sum ne^{(-n^2)}
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  6. #6
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    \sum \dfrac{1}{n\ln(n)} \text{converges} \Leftrightarrow \int \dfrac{dx}{x\ln(x)} \text{converges}

    Using the substitution u=ln(x)

    \int \frac{du}{u}=\ln(\ln(x))

    Evaluating at infinity, the integral diverges. Then, the serie diverges also.
    Last edited by Krizalid; February 4th 2010 at 07:58 AM.
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  7. #7
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    felper: thanks for reply but its an a-level text (secondary school 16-18 year olds )and i dont have that rule at my disposal. Only ratio/comparison test. I think as they give an answer of convergent as the answer, the typo is with the question and not the answer.

    Ive googled and can not find an obvious comparison to use with this sum,all i can find are tests that A-level students would not be expected to know as part of the course.

    Maybe the question was \sum \frac {1}{n^{b}ln(n)} with b>1,which i can do with a comparison.

    can i have a confirmation that my answer of convergent is correct for
    \sum ne^{(-n^2)}

    Finally,i have been moved. Did i post in wrong forum?
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