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  1. #1
    413
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    Series

    For what values of p>0 does the series

    a) Riemann Sum [n=1 to infinity] 1/ [n(ln n)^p]

    converge and for what values does it diverge?

    I am having trouble doing this question, please help.
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  2. #2
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    The infinite series,
    \sum_{n=2}^{\infty} \frac{1}{n^a \ln^b n}
    Is slightly famous.
    (Called Bertrand Series)

    The results says that if,
    a>1---> converges
    a<1---> diverges
    a=1---> depends.

    The value of b makes the dependance.
    If a=1 then it converges only if b>1.

    Therefore, your series converges when p>1
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  3. #3
    413
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    but i didn't learn the Bertrand Series, is there another approach to this question?
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  4. #4
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    The function,
    f(x)=\frac{1}{x\ln^ p x}
    Is a positive, decreasing, continous function for x>1
    With the property that,
    f(n)=a_n in the series.

    You can apply the integral series test.
    And see when, (I am just goint to change the lower limit, it makes no difference for convergence).
    \int_2^{\infty} \frac{1}{x\ln^ p x}dx
    If,
    u=\ln x then, u'=1/x
    Thus, by the substitution rule,
    \int \frac{1}{x\ln^p x}=\int u^{-p} du
    Since p>0 the antiderivative depends whether p is 1 or not.
    So we will divide this problem into 3 cases,
    0<p<1
    p=1
    p>1

    It, 0<p<1 then,
    \int u^{-p}du=\frac{u^{1-p}}{1-p}+C
    Thus, (dropping the last term, makes no difference)
    \int_2^{\infty} \frac{x\ln^p x}dx=\lim_{n\to\infty} \frac{\ln^{1-p} x}{1-p}
    Which will increase without bound for,
    1-p>0

    If p=1 then,
    \int u^{-p}du=\ln u+C
    Thus, (dropping the last term, makes no difference)
    \int_2^{\infty} \frac{x}{\ln x}dx=\lim_{x\to\infty} \ln ln x.
    Which will increase without bound for,
    \ln \ln x is increasing (though very very very, snail slowly).

    If p>1 then,
    \int u^{-p}du=\frac{u^{-p+1}}{-p+1}+C
    Thus, (dropping the last term, makes no difference)
    \int_2^{\infty} \frac{x}{\ln^p x}=\lim_{x\to \infty} \frac{\ln ^{-p+1} x}{-p+1}
    Which will converge for, -p+1<0 and this assures us the fraction is in the denominator with the denominator increasing without bound, so the limit converges.
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  5. #5
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    another question asking for the same thing is this...

    Riemann Sum [n=1 to infinity] 1/ [n(ln n) (ln(ln n))^p]

    do you still use the same method? but this one seems so hard to integrate, please help. thanks.
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    Quote Originally Posted by 413 View Post
    another question asking for the same thing is this...

    Riemann Sum [n=1 to infinity] 1/ [n(ln n) (ln(ln n))^p]

    do you still use the same method? but this one seems so hard to integrate, please help. thanks.
    The the solution is based on,
    \int \frac{1}{x\ln x \ln (\ln^p x)}dx
    First for x\geq 3 and p>0 (note the reason why I make these inequalities is because otherwise the function might not be defined or I will not be able to get integrals).
    Then, \ln (\ln^p x)=p\ln \ln x
    Thus, the integral becomes,
    \frac{1}{p}\int \frac{1}{x\ln x\ln \ln x}
    Express this integral as,
    \int \frac{1}{\ln x\ln \ln x}\cdot \frac{1}{x} dx
    If, u=\ln x then u'=1/x
    Thus,
    \int \frac{1}{u\ln u}\cdot u' dx
    Substitution rule,
    \int \frac{1}{u\ln u} du
    Express as,
    \int \frac{1}{\ln u}\cdot \frac{1}{u}
    Let,
    t=\ln u then, t'=1/u thus,
    \int \frac{1}{t} t'du
    By the substitution rule we have,
    \int \frac{1}{t} dt=\ln |t|+C
    Thus,
    \ln |\ln u|+C
    Thus,
    \ln |\ln \ln x|+C is the integral.
    Thus, the original integral is,
    \frac{1}{p}\ln |\ln \ln x|+C
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  7. #7
    413
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    but the exponent p is outside the second ln, does that make a difference?

    1/ [n(ln n) (ln(ln n))^p]
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  8. #8
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    is the answer divergent for all p?
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  9. #9
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    Quote Originally Posted by 413 View Post
    but the exponent p is outside the second ln, does that make a difference?

    1/ [n(ln n) (ln(ln n))^p]
    \int \frac{1}{x\ln x (\ln \ln x)^p} dx

    Let,
    u=\ln x then, u'=1/x
    Thus,
    \int \frac{1}{u (\ln u)^p}du
    Let,
    t=\ln u then, t'=1/u
    Thus,
    \int \frac{1}{t^p}dt=\int t^{-p}dt=\left\{ \begin{array}{c}\frac{t^{1-p}}{1-p}+C\mbox{ if }p\not = 1\\ \ln t +C\mbox{ if }p=1 \end{array} \right\}
    That is how you integrate it.

    Double substitution,
    \left\{ \begin{array}{c}\frac{(\ln \ln x)^{1-p}}{1-p}+C\mbox{ if }p\not = 1\\ \ln \ln \ln x+C \mbox{ if }p=1 \end{array} \right\}
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  10. #10
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    it is still divergent for all p right?
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  11. #11
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    Quote Originally Posted by 413 View Post
    it is still divergent for all p right?
    How! Did you try to solve it? No.

    Look at the integral. For the same reason it will diverge for p=1 for \ln \ln \ln x is divergent.

    And if the exponent 1-p>0 then you have again an increasing function so it is divergent. When 1-p<0 then the exponent is negative thus the function will eventually die. In that case is convergese for p>1
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