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Math Help - Series (Convergence, Divergence)

  1. #1
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    Series (Convergence, Divergence)

    Was doing my homework, came up with 5 I could not get.

    I'm not too knowledgeable about the special characters you guys use to make the math problems, so I'll do my best to use normal characters.. (keep in mind that each of these problems starts with E (sum), and unless otherwise noted, it starts at n=0 and goes to infinity)

    #1: (x+3)^n / (2)^n, converge or diverge, and find the sum if it converges

    I was thinking of using the Geometric test, but I don't know how that works when there is an x involved (1/3 converges, but does ((1+x)/3) converge?)

    #2: 1/5 + 1/8 + 1/11 + 1/14 + 1/17

    I wrote up the sum (1/(5 + 3n)) where n=0, goes to infinity

    I imagine then that using the p-test, it diverges? or is this not valid?

    #3 asks what value of p would make the series converge?

    series is (n=1, to infinity): ln(p) / n^p

    I'm pretty sure you need the integral test, but how do you take the integral of ln(x)/x^p?

    #4: 1+ sin(n) / 10^n

    Comparison test I'm guessing? But what do you compare it to? And If you do the inequality bit (0<1+sin(n)<2), how do you determine if 2/10^n converges?

    #5 (n^2 - 5n) / (n^3 + n + 1)

    Can you just use L'hopital? (2n - 5) / (2n^3 +1), which becomes 2 / 6n, which diverges?

    As you can tell, I am not too good with these..

    I tried to make it as clear as possible, but if you're confused about the way I conveyed a question, I'll gladly try to clarify further
    Last edited by coolguy99; March 15th 2009 at 08:51 PM.
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  2. #2
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    Quote Originally Posted by coolguy99 View Post
    Was doing my homework, came up with 5 I could not get.

    I'm not too knowledgeable about the special characters you guys use to make the math problems, so I'll do my best to use normal characters.. (keep in mind that each of these problems starts with E (sum), and unless otherwise noted, it starts at n=0 and goes to infinity)

    #1: (x+3)^n / (2)^n, converge or diverge, and find the sum if it converges

    I was thinking of using the Geometric test, but I don't know how that works when there is an x involved (1/3 converges, but does ((1+x)/3) converge?)

    #2: 1/5 + 1/8 + 1/11 + 1/14 + 1/17

    I wrote up the sum (1/(5 + 3n)) where n=0, goes to infinity

    I imagine then that using the p-test, it diverges? or is this not valid?

    #3 asks what value of p would make the series converge?

    series is (n=1, to infinity): ln(p) / n^p

    I'm pretty sure you need the integral test, but how do you take the integral of ln(x)/x^p?

    #4: 1+ sin(n) / 10^n

    Comparison test I'm guessing? But what do you compare it to? And If you do the inequality bit (0<1+sin(n)<2), how do you determine if 2/10^n converges?

    #5 (n^2 - 5n) / (n^3 + n + 1)

    Can you just use L'hopital? (2n - 5) / (2n^3 +1), which becomes 2 / 6n, which diverges?

    As you can tell, I am not too good with these..

    I tried to make it as clear as possible, but if you're confused about the way I conveyed a question, I'll gladly try to clarify further
    #1:  \sum_{n=1}^\infty \left( \frac{x+3}{2} \right)^n converges if  \left| \frac{x+3}{2} \right| < 1. If so, the sum is S = \frac{a}{1-r} where a is the first term and r =\frac{x+3}{2}

    #2  \sum_{n=1}^\infty \frac{1}{3n+5} LCT with  \sum_{n=0}^\infty \frac{1}{n}

    #3  \int \frac{\ln x}{x^p}\,dx use parts

    #4:  \sum_{n=1}^\infty \frac{1+\sin n}{10^n} \le \sum_{n=1}^\infty \frac{2}{10^n} and use the direct comparison test.

    #5  \sum_{n=1}^\infty \frac{n^2-5n}{n^3+n+1} LCT with  \sum_{n=1}^\infty \frac{1}{n}
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    #1:  \sum_{n=1}^\infty \left( \frac{x+3}{2} \right)^n converges if  \left| \frac{x+3}{2} \right| < 1. If so, the sum is S = \frac{a}{1-r} where a is the first term and r =\frac{x+3}{2}
    I'm getting... (x+3) / (5-x) as a sum right now... assuming that a = (x+3)/(2) and r = (x+3)/(2^n-1)

    Quote Originally Posted by danny arrigo View Post
    #2  \sum_{n=1}^\infty \frac{1}{3n+5} LCT with  \sum_{n=0}^\infty \frac{1}{n}
    I got diverges, which I think is right, thanks

    Quote Originally Posted by danny arrigo View Post
    #3  \int \frac{\ln x}{x^p}\,dx use parts
    I'm still having trouble integrating..
    u = lnx
    du = 1/x dx
    dv = 1/(x^p)
    v = (p+1)/(x^p+1)

    ?
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  4. #4
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    Quote Originally Posted by coolguy99 View Post
    I'm getting... (x+3) / (5-x) as a sum right now... assuming that a = (x+3)/(2) and r = (x+3)/(2^n-1)



    I got diverges, which I think is right, thanks



    I'm still having trouble integrating..
    u = lnx
    du = 1/x dx
    dv = 1/(x^p)
    v = (p+1)/(x^p+1)

    ?
    Write your dv as dv = x^{-p} and take p = 1 as a special case.
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  5. #5
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    Alright, I see what you're getting at with that one... But how do you integrate vdu?

    And about that first one,

    I split it up into ((x+3)/2) x ((x+3)/2))^(n-1)


    I have the sum = (x+3) / (5-x), but I'm guessing that is incomplete?
    Last edited by coolguy99; March 16th 2009 at 07:35 AM.
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  6. #6
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    Quote Originally Posted by coolguy99 View Post
    Alright, I see what you're getting at with that one... But how do you integrate vdu?

    And about that first one,

    I split it up into ((x+3)/2) x ((x+3)/2))^(n-1)


    I have the sum = (x+3) / (5-x), but I'm guessing that is incomplete?
    Not sure how you got that sum \frac{\frac{x+3}{2}}{1 - \frac{x+3}{2}} = - \frac{x+3}{x+1}. As for your integration question

    u = \ln x,\; dv = x^{-p}dx or
    du = \frac{dx}{x},\; v =\frac{x^{1-p}}{1-p}

    so
    \int vdu = uv - \int u dv = \frac{x^{1-p} \ln x }{1-p} - \frac{1}{1-p} \int x^{1-p}\frac{dx}{x} = \frac{x^{1-p} \ln x }{1-p} - \frac{1}{1-p} \int x^{-p}\,dx = \frac{x^{1-p} \ln x }{1-p} - \frac{1}{(1-p)^2} x^{1-p}. Again, p = 1 is a special case.
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  7. #7
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    Alright, so that makes sense..

    The question originally asked at what value for P would make the series convergent.

    so you do the lim (as x->infinity) of
     \frac{x^{1-p} \ln x }{1-p} - \frac{1}{(1-p)^2} x^{1-p}<br />

    on the interval... (1, infinity)?

    But I'm not quite sure how you would interpret that
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