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Math Help - Test comparison

  1. #1
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    Test comparison

    Resolving a list of exercise these two exercises I could not

    Determined using the comparison test

    1) \sum_{n=1}^{\infty} \frac{(n+1)}{n^3}

    2) \sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Resolving a list of exercise these two exercises I could not

    Determined using the comparison test

    1) \sum_{n=1}^{\infty} \frac{(n+1)}{n^3}

    2) \sum_{n=0}^{\infty} \frac{n^2+1}{n^3+1}
    The limit comparison test says that:

    • If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges.
    • If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
    • If the limit of a[n]/b[n] is infinite, and the sum of b[n] diverges, then the sum of a[n] also diverges.

    For the first one, compare it to \frac{1}{n^2}. \lim_{n\to\infty}\frac{\frac{(n+1)}{n^3}}{\frac{1}  {n^2}} = 1, so the sum converges because \sum_{n=1}^{\infty}\frac{1}{n^2} does.

    For the second one, compare it to \frac{1}{n+1}. \lim_{n\to\infty}\frac{\frac{n^2+1}{n^3+1}}{\frac{  1}{n+1}}=1, so the sum diverges because \sum_{n=0}^{\infty}\frac{1}{n+1} does.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    The limit comparison test says that:

    • If the limit of a[n]/b[n] is positive, then the sum of a[n] converges if and only if the sum of b[n] converges.
    • If the limit of a[n]/b[n] is zero, and the sum of b[n] converges, then the sum of a[n] also converges.
    • If the limit of a[n]/b[n] is infinite, and the sum of b[n] diverges, then the sum of a[n] also diverges.

    For the first one, compare it to \frac{1}{n^2}. \lim_{n\to\infty}\frac{\frac{(n+1)}{n^3}}{\frac{1}  {n^2}} = 1, so the sum converges because \sum_{n=1}^{\infty}\frac{1}{n^2} does.

    For the second one, compare it to \frac{1}{n+1}. \lim_{n\to\infty}\frac{\frac{n^2+1}{n^3+1}}{\frac{  1}{n+1}}=1, so the sum diverges because \sum_{n=0}^{\infty}\frac{1}{n+1} does.

    Thank you very much
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  4. #4
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    i read comparison test, so we can actually use the direct comparison test.

    1) \frac{n+1}{n^{3}}\le \frac{2n}{n^{3}}=\frac{2}{n^{2}}.

    2) \frac{n^{2}+1}{n^{3}+1}>\frac{n^{2}}{2n^{3}}=\frac  {1}{2n}.

    As long as n\ge1.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    i read comparison test, so we can actually use the direct comparison test.

    1) \frac{n+1}{n^{3}}\le \frac{2n}{n^{3}}=\frac{2}{n^{2}}.

    2) \frac{n^{2}+1}{n^{3}+1}>\frac{n^{2}}{2n^{3}}=\frac  {1}{2n}.

    As long as n\ge1.
    I have a doubt.
    If I have one series a_n I want to determine the convergence, found that the series b_n is convergent, but b_n > a_n up to a value of n, not for all values of n. The series a_n is convergent ?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    I have a doubt.
    If I have one series a_n I want to determine the convergence, found that the series b_n is convergent, but b_n > a_n up to a value of n, not for all values of n. The series a_n is convergent ?
    Not necessarily. If you are comparing a_n to b_n and you know that b_n converges, a_n\leq b_n for all but finitely many n will tell you that a_n converges.

    For example, say b_n converges. You compare a_n to b_n and find:

    1.) \left\{\begin{array}{lr}a_n<b_n &: n\leq 1000\\a_n>b_n &: n>1000\end{array}\right\} a_n does not necessarily converge.

    2.) \left\{\begin{array}{lr}a_n>b_n &: n\leq 1000\\a_n<b_n &: n>1000\end{array}\right\} a_n will converge.

    Note that my choice of 1000 was arbitrary.
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  7. #7
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    Quote Originally Posted by redsoxfan325 View Post
    Not necessarily. If you are comparing a_n to b_n and you know that b_n converges, a_n\leq b_n for all but finitely many n will tell you that a_n converges.

    For example, say b_n converges. You compare a_n to b_n and find:

    1.) \left\{\begin{array}{lr}a_n<b_n &: n\leq 1000\\a_n>b_n &: n>1000\end{array}\right\} a_n does not necessarily converge.

    2.) \left\{\begin{array}{lr}a_n>b_n &: n\leq 1000\\a_n<b_n &: n>1000\end{array}\right\} a_n will converge.

    Note that my choice of 1000 was arbitrary.

    If not for all but finitely many I can not say that converge?
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  8. #8
    Super Member redsoxfan325's Avatar
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    No. It could possibly converge, but the comparison test does not prove it.
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  9. #9
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    Quote Originally Posted by redsoxfan325 View Post
    No. It could possibly converge, but the comparison test does not prove it.
    Thank you.


    This series converge?


    \sum_{n=2}^{\infty} \frac{1}{n\sqrt{n^2-1}}
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  10. #10
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Thank you.


    This series converge?


    \sum_{n=2}^{\infty} \frac{1}{n\sqrt{n^2-1}}
    Try the limit comparison test with b_n=\frac{1}{n^2}.

    \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = ???
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  11. #11
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    Quote Originally Posted by redsoxfan325 View Post
    Try the limit comparison test with b_n=\frac{1}{n^2}.

    \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = ???
    This limit is not a number
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  12. #12
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    This limit is not a number
    \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1

    So a_n converges because b_n does.
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  13. #13
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    Quote Originally Posted by redsoxfan325 View Post
    \lim_{n\to\infty}\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}} = \lim_{n\to\infty}\frac{n}{n\sqrt{1-1/n^2}} = \lim_{n\to\infty}\frac{1}{\sqrt{1-1/n^2}}=\frac{1}{\sqrt{1-0}}=1

    So a_n converges because b_n does.
    Thank you
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