Results 1 to 11 of 11

Math Help - Series

  1. #1
    413
    413 is offline
    Junior Member
    Joined
    Oct 2006
    Posts
    32

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    413
    413 is offline
    Junior Member
    Joined
    Oct 2006
    Posts
    32
    but i didn't learn the Bertrand Series, is there another approach to this question?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    413
    413 is offline
    Junior Member
    Joined
    Oct 2006
    Posts
    32
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    413
    413 is offline
    Junior Member
    Joined
    Oct 2006
    Posts
    32
    but the exponent p is outside the second ln, does that make a difference?

    1/ [n(ln n) (ln(ln n))^p]
    Follow Math Help Forum on Facebook and Google+

  8. #8
    413
    413 is offline
    Junior Member
    Joined
    Oct 2006
    Posts
    32
    is the answer divergent for all p?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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\}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    413
    413 is offline
    Junior Member
    Joined
    Oct 2006
    Posts
    32
    it is still divergent for all p right?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 3rd 2011, 01:12 AM
  2. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  3. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  4. Replies: 2
    Last Post: December 1st 2009, 12:45 PM
  5. Replies: 1
    Last Post: May 5th 2008, 09:44 PM

Search Tags


/mathhelpforum @mathhelpforum