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Math Help - Test for convergence

  1. #1
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    Question Test for convergence

    For the following summations from 1 to infinity, which tests for convergence would I use? I am confused on what to select.

    1) 9/((n^2)*(sqrt(n)))

    2) (-3*n*sqrt(n))/(n^2)

    3) 3*(-1)^n

    4) cos((n^5)+1)

    5) (n^7)-10*n

    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by taurus View Post
    For the following summations from 1 to infinity, which tests for convergence would I use? I am confused on what to select.

    1) 9/((n^2)*(sqrt(n)))

    2) (-3*n*sqrt(n))/(n^2)

    3) 3*(-1)^n

    4) cos((n^5)+1)

    5) (n^7)-10*n

    Thanks
    1) \sum_{n=1}^{\infty}\frac{9}{n^2\sqrt{n}}\implies \sum_{n=1}^{\infty}\frac{9}{n^{\frac{5}{2}}}. You can apply the integral test, or recognize that it's a p-series, where \frac{5}{2}>1 (thus, the series converges).

    2) \sum_{n=1}^{\infty}\frac{-3n\sqrt{n}}{n^2} \implies \sum_{n=1}^{\infty}-\frac{3}{n^{\frac{1}{2}}}. Again apply the integral test, or recognize it's a p-series, where \frac{1}{2}<1 (thus, the series diverges).

    3) I forget how to prove it, but it diverges. (expanding we get 3-3+3-3+3-3+...)

    4) \sum_{n=1}^{\infty}\cos(n^5+1). I believe you would use the limit comparison test here (?) : Thus, \lim_{n\to{\infty}}cos(n^5) \ \text{DNE}. Since this doesn't exist, then the series diverges. (I think this is correct)

    5) \sum_{n=1}^{\infty}n^7-10n. I'm sure the integral test would work here as well. It's pretty evident that the series diverges.

    Can someone verify these??
    Last edited by Chris L T521; May 23rd 2008 at 11:47 PM. Reason: Syntax errors
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Chris L T521 View Post
    3) I forget how to prove it, but it diverges. (expanding we get 3-3+3-3+3-3+...)

    4) \sum_{n=1}^{\infty}\cos(n^5+1). I believe you would use the limit comparison test here (?) : Thus, \lim_{n\to{\infty}}cos(n^5) \ \text{DNE}. Since this doesn't exist, then the series diverges. (I think this is correct)

    5) \sum_{n=1}^{\infty}n^7-10n. I'm sure the integral test would work here as well. It's pretty evident that the series diverges.

    Can someone verify these??
    For the third and the fifth series, you can use the same method as for the fourth one : if \lim_{n\to \infty} a_n \neq 0 than the series \sum a_n diverges.
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  4. #4
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    Question

    so for
    n^7 - 10n
    using the limit comparison test, does that mean i can just put a few numbers in as 'n' and if it doenst appear close to any number it means it diverges?
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by taurus View Post
    so for
    n^7 - 10n
    using the limit comparison test, does that mean i can just put a few numbers in as 'n' and if it doenst appear close to any number it means it diverges?
    Yes, you can, for example, say that " \lim_{n\to\infty}n^7-10n=\infty\neq 0 hence \sum n^7-10n diverges."
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  6. #6
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    Question

    oh so if i put any number as i get bigger it has to go close to zero?

    also for the integral test, after you integrate, what value to you use as upper bound?

    like for example this says n^2 + 7:
    Visual Calculus -

    i dont quite get the step after they say the integral becomes....


    EDIT: ahh alright i get it
    Last edited by taurus; May 24th 2008 at 12:41 AM.
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  7. #7
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    i know 1/(7n+4) doesnt converge but how would i prove it?
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  8. #8
    Moo
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    Quote Originally Posted by taurus View Post
    i know 1/(7n+4) doesnt converge but how would i prove it?
    Because past a certain value of n, \frac{1}{7n+4} \sim \frac{1}{7n}

    And \sum \frac{1}{7n}=\frac 17 \sum \frac 1n, which diverges
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  9. #9
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    where did the 4 go?
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  10. #10
    Moo
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    Quote Originally Posted by taurus View Post
    where did the 4 go?
    When n becomes very big, 4 is being very small comparing to 7n. So we can neglect him.

    If you prefer : \lim_{n \to \infty} 7n+4=\lim_{n \to \infty} 7n
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  11. #11
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    Question

    So for this:
    (2n+5)/((n+2)^4)


    what rule would i use?

    Would it be like this:
    > (2n+5)/((n+2)^4)
    > 2n/(n+2)^4
    > 2n/n^4
    > 2/n^3
    so p series with p > 1 therefore it converges?
    Last edited by taurus; May 24th 2008 at 02:39 AM.
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  12. #12
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by taurus View Post
    So for this:
    (2n+5)/((n+2)^4)


    what rule would i use?

    Would it be like this:
    > (2n+5)/((n+2)^4)
    > 2n/(n+2)^4
    > 2n/n^4
    > 2/n^3
    so p series with p > 1 therefore it converges?
    That's it !
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  13. #13
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    what about
    (3^n)/(4+(8^n))

    i got this far:
    (3^n)/(8^n)

    but from there what?
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  14. #14
    Moo
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    Quote Originally Posted by taurus View Post
    what about
    (3^n)/(4+(8^n))

    i got this far:
    (3^n)/(8^n)

    but from there what?
    \frac{3^n}{8^n}=\left(\frac 38\right)^n

    This is the general term for a geometric sequence.

    \sum_{n=0}^\infty \left(\frac 38\right)^n=\lim_{n \to \infty} \frac{1-\left(\frac 38\right)^n}{1-\frac 38}

    It converges because \frac 38 <1 \implies \lim_{n \to \infty} \left(\frac 38\right)^n=0
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  15. #15
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    Question

    Quote Originally Posted by Moo View Post
    Because past a certain value of n, \frac{1}{7n+4} \sim \frac{1}{7n}

    And \sum \frac{1}{7n}=\frac 17 \sum \frac 1n, which diverges

    But doesnt my number go to 0 as n becomes larger? therefore its convergent?
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