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Math Help - [SOLVED] Infinite series, convergent or not?

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Infinite series, convergent or not?

    It has probably been answered before on MHF, but I didn't find it using the search button.
    Tell whether the following infinite series is convergent or not :
    \sum_{n=2}^{+\infty} \frac{1}{n \ln (n)}. At first, I used the ration test, but I cannot conclude since it says "1". Then the integral test, which gave me divergent, but I'm not sure I did it well, because when I tried to get an approximation via Mathematica, I got an error like if it converges, it does it too slow or something like that. Nevertheless, it said an approximation is 13.4391. From it I doubted my integral test was well made. I tried the root test, but I'm not sure how to finish it.
    Integral test : \int_2^{+\infty} \frac{1}{x\ln (x)}. Let u be \ln (x), then du=\frac{1}{x}, and the integral becomes \int_2^{+\infty} \frac{du}{u}=\ln u \big | ^{+\infty}_2. Replacing u by \ln (u), it's easy to see the integral diverges.
    So normally, so does the infinite series. Am I right?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    It has probably been answered before on MHF, but I didn't find it using the search button.
    Tell whether the following infinite series is convergent or not :
    \sum_{n=2}^{+\infty} \frac{1}{n \ln (n)}. At first, I used the ration test, but I cannot conclude since it says "1". Then the integral test, which gave me divergent, but I'm not sure I did it well, because when I tried to get an approximation via Mathematica, I got an error like if it converges, it does it too slow or something like that. Nevertheless, it said an approximation is 13.4391. From it I doubted my integral test was well made. I tried the root test, but I'm not sure how to finish it.
    Integral test : \int_2^{+\infty} \frac{1}{x\ln (x)}. Let u be \ln (x), then du=\frac{1}{x}, and the integral becomes \int_2^{+\infty} \frac{du}{u}=\ln u \big | ^{+\infty}_2. Replacing u by \ln (u), it's easy to see the integral diverges.
    So normally, so does the infinite series. Am I right?
    in terms of u, the limits are \int_{\ln 2}^{\infty} \frac{du}{u}=\ln u \bigg | ^{\infty}_{\ln 2}

    yes, it is divergent
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    It has probably been answered before on MHF, but I didn't find it using the search button.
    Tell whether the following infinite series is convergent or not :
    \sum_{n=2}^{+\infty} \frac{1}{n \ln (n)}. At first, I used the ration test, but I cannot conclude since it says "1". Then the integral test, which gave me divergent, but I'm not sure I did it well, because when I tried to get an approximation via Mathematica, I got an error like if it converges, it does it too slow or something like that. Nevertheless, it said an approximation is 13.4391. From it I doubted my integral test was well made. I tried the root test, but I'm not sure how to finish it.
    Integral test : \int_2^{+\infty} \frac{1}{x\ln (x)}. Let u be \ln (x), then du=\frac{1}{x}, and the integral becomes \int_2^{+\infty} \frac{du}{u}=\ln u \big | ^{+\infty}_2. Replacing u by \ln (u), it's easy to see the integral diverges.
    So normally, so does the infinite series. Am I right?

    \int_2^{\infty}\frac{dx}{x\ln^p(x)}


    Now lets look at the five cases

    Case one

    If p<0

    \int_2^{\infty}\frac{dx}{x\ln^p(x)} can be rewritten as \int_2^{\infty}\frac{\ln^{-p}(x)}{x}dx where -p>0\quad\text{because }p<0

    So then

    \int_2^{\infty}\frac{\ln^{-p}(x)}{x}dx>\int_2^{\infty}\frac{dx}{x}

    Therefore divergent.

    Now what about when p=0

    Obviously then we have

    \int\frac{dx}{x\ln^0(x)}=\int_1^{\infty}\frac{dx}{  x}=\infty

    So divergent.

    How about when

    0<p<1

    We then have that

    \int\frac{dx}{x\ln^p(x)}dx=\frac{\ln^{-p+1}(x)}{-p+1}

    Now at x=2\Rightarrow\frac{\ln^{-p+1}(2)}{-p+1}\text{ Which is a finite value}

    Now for x\to\infty consider that since 0<p<1\Rightarrow{-p+1>0}

    And it can can be shown by direct substitution that \lim_{x\to\infty}\ln^{n\in(0,1)}(x)=\infty

    Therefore divergent.


    Now when p=1 we have that

    \int_2^{\infty}\frac{dx}{x\ln(x)}=\ln(\ln(x))\bigg  |_2^{\infty}=\infty-\ln(\ln(2))

    So divergent.

    Now how about when

    p>1

    we have that

    \int_2^{\infty}\frac{dx}{x\ln^p}=\frac{\ln^{-p+1}(x)}{-p+1}\bigg|_{2}^{\infty}

    Now consider that

    p>1\Rightarrow{-p+1<0}

    Now it can be shown very easily that \lim_{x\to\infty}\ln^{n\in(-\infty,0)}(x)=0

    So then

    \int_2^{\infty}\frac{dx}{x\ln^p(x)}=0-\frac{\ln^{-p+1}(2)}{-p+1}

    Therefore convergent.


    Putting this together we get

    \sum_{n=2}^{\infty}\frac{1}{n\ln^p(n)}

    Converges only for p>1
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