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Math Help - How to show this sequence converges?

  1. #1
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    How to show this sequence converges?

    Show that
    a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1) converges to some real number.

    The problem hints that I'm supposed to use the fact that
    \sum_{k=1}^\infty \frac{x}{k(x+k)} converges uniformly on [0,1], which I've already proved.

    I can't figure out how to connect these two facts at all... any help?
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  2. #2
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    Quote Originally Posted by paupsers View Post
    Show that
    a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1) converges to some real number.

    The problem hints that I'm supposed to use the fact that
    \sum_{k=1}^\infty \frac{x}{k(x+k)} converges uniformly on [0,1], which I've already proved.

    I can't figure out how to connect these two facts at all... any help?

    My opinion :

    Can we use that


     \frac{1}{k} > \ln(1 + \frac{1}{k}) > \frac{1}{k+1} ?


    then the series  \left( \sum_{k=1}^n \frac{1}{k} \right ) - \ln(n+1)

     = \left( \sum_{k=1}^n \frac{1}{k} \right ) - \sum_{k=1}^n \ln( 1 + \frac{1}{k} )

     = \sum_{k=1}^n \left(  \frac{1}{k} - \ln( 1 + \frac{1}{k}  ) \right) < \sum_{k=1}^n \left(  \frac{1}{k} - \frac{1}{k+1} \right)    = \sum_{k=1}^n \frac{1}{k(k+1)} which is convergent .


    Also , since  \frac{1}{k} > \ln(1 + \frac{1}{k} ) , the series is greater than zero , it is bounded .

    I think there may be some errors but anyway , you can take a look and perhaps it can help you a little .
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  3. #3
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    Quote Originally Posted by paupsers View Post
    Show that
    a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1) converges to some real number.

    The problem hints that I'm supposed to use the fact that
    \sum_{k=1}^\infty \frac{x}{k(x+k)} converges uniformly on [0,1], which I've already proved.

    I can't figure out how to connect these two facts at all... any help?
    Let's use your hint.

    Because of the uniform convergence, the sum f(x)=\sum_{k=1}^\infty \frac{x}{k(x+k)} is continuous on [0,1]. Furthermore, you may integrate term by term on [0,1]. Note also that \frac{x}{k(x+k)}=\frac{1}{k}-\frac{1}{x+k}. We get:

    \sum_{k=1}^\infty \left(\frac{1}{k}-(\log(k+1)-\log k)\right)=\int_0^1 f(x) dx.

    The important thing is that the right-hand side is finite.

    The above shows (means) that \sum_{k=1}^n \left(\frac{1}{k}-(\log(k+1)-\log k)\right) has a finite limit as n\to\infty.

    However, this partial sum can be rewritten as \left(\sum_{k=1}^n \frac{1}{k}\right)-\log (n+1)...
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  4. #4
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    Quote Originally Posted by Laurent View Post
    We get:

    \sum_{k=1}^\infty \left(\frac{1}{k}-(\log(k+1)-\log k)\right)=\int_0^1 f(x) dx.
    I don't understand how you got to this point. What exactly did you integrate?
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  5. #5
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    Quote Originally Posted by paupsers View Post
    I don't understand how you got to this point. What exactly did you integrate?
    f between 0 and 1...

    \int_0^1 f(x)dx = \int_0^1 \left(\sum_{k=1}^\infty \frac{x}{k(x+k)}\right)dx =\sum_{k=1}^\infty \int_0^1 \frac{x}{k(x+k)}dx,

    (last step justified by uniform convergence) and used the expression I gave to simplify the computation of the last integral.
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