Asymptotic formula for $sum_{nle x}frac 1n$












2












$begingroup$



Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory







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$endgroup$












  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56


















2












$begingroup$



Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56
















2












2








2


1



$begingroup$



Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory







share|cite|improve this question











$endgroup$





Prove that $displaystyle sum_{nle x}frac 1n=log (x+2)+O(1)$.




We have, $displaystyle sum_{nle x}frac 1n=log x +gamma+O(1/x)$. Now if I can show that $log(x+2)-log x+O(1/x)=O(1)$ then I have done.



Now, $log(x+2)-log x+O(1/x)=log left(frac{x+2}{x}right)+O(1/x)=O(1/x).$



How can I prove this ?






complex-analysis number-theory analytic-number-theory




complex-analysis number-theory analytic-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 19:29







Topo

















asked Dec 25 '18 at 4:59









TopoTopo

330314




330314












  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56




















  • $begingroup$
    I don't understand the thing in yellow. I've never seen all big Os in an equation
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:35










  • $begingroup$
    @mathworker21 Fixed.
    $endgroup$
    – Topo
    Dec 25 '18 at 5:50










  • $begingroup$
    Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
    $endgroup$
    – mathworker21
    Dec 25 '18 at 5:56


















$begingroup$
I don't understand the thing in yellow. I've never seen all big Os in an equation
$endgroup$
– mathworker21
Dec 25 '18 at 5:35




$begingroup$
I don't understand the thing in yellow. I've never seen all big Os in an equation
$endgroup$
– mathworker21
Dec 25 '18 at 5:35












$begingroup$
@mathworker21 Fixed.
$endgroup$
– Topo
Dec 25 '18 at 5:50




$begingroup$
@mathworker21 Fixed.
$endgroup$
– Topo
Dec 25 '18 at 5:50












$begingroup$
Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
$endgroup$
– mathworker21
Dec 25 '18 at 5:56






$begingroup$
Also, you don't need the $O(1)$ if you have the $O(log (x+2))$
$endgroup$
– mathworker21
Dec 25 '18 at 5:56












1 Answer
1






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oldest

votes


















4












$begingroup$

All you need is
$log(1+z) < z$
for $z > 0$.



Then
$log(x+2)-log(x)
=log(1+frac{2}{x})
lt frac{2}{x}
=O(frac1{x})
$.



To show
$log(1+z) < z$
for $z > 0$:



$log(1+z)
=int_0^z frac{dt}{1+t}
lt int_0^z dt
=z
$.



You can also use the integral
to get a lower bound:



$begin{array}\
log(1+z)
&=int_0^z frac{dt}{1+t}\
&gt int_0^z frac{dt}{1+z}\
&=frac{z}{1+z}\
&=frac{z+z^2-z^2}{1+z}\
&=z-frac{z^2}{1+z}\
&gt z-z^2\
end{array}
$






share|cite|improve this answer









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    1 Answer
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    1






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    4












    $begingroup$

    All you need is
    $log(1+z) < z$
    for $z > 0$.



    Then
    $log(x+2)-log(x)
    =log(1+frac{2}{x})
    lt frac{2}{x}
    =O(frac1{x})
    $    .



    To show
    $log(1+z) < z$
    for $z > 0$:



    $log(1+z)
    =int_0^z frac{dt}{1+t}
    lt int_0^z dt
    =z
    $    .



    You can also use the integral
    to get a lower bound:



    $begin{array}\
    log(1+z)
    &=int_0^z frac{dt}{1+t}\
    &gt int_0^z frac{dt}{1+z}\
    &=frac{z}{1+z}\
    &=frac{z+z^2-z^2}{1+z}\
    &=z-frac{z^2}{1+z}\
    &gt z-z^2\
    end{array}
    $






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      All you need is
      $log(1+z) < z$
      for $z > 0$.



      Then
      $log(x+2)-log(x)
      =log(1+frac{2}{x})
      lt frac{2}{x}
      =O(frac1{x})
      $      .



      To show
      $log(1+z) < z$
      for $z > 0$:



      $log(1+z)
      =int_0^z frac{dt}{1+t}
      lt int_0^z dt
      =z
      $      .



      You can also use the integral
      to get a lower bound:



      $begin{array}\
      log(1+z)
      &=int_0^z frac{dt}{1+t}\
      &gt int_0^z frac{dt}{1+z}\
      &=frac{z}{1+z}\
      &=frac{z+z^2-z^2}{1+z}\
      &=z-frac{z^2}{1+z}\
      &gt z-z^2\
      end{array}
      $






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        All you need is
        $log(1+z) < z$
        for $z > 0$.



        Then
        $log(x+2)-log(x)
        =log(1+frac{2}{x})
        lt frac{2}{x}
        =O(frac1{x})
        $        .



        To show
        $log(1+z) < z$
        for $z > 0$:



        $log(1+z)
        =int_0^z frac{dt}{1+t}
        lt int_0^z dt
        =z
        $        .



        You can also use the integral
        to get a lower bound:



        $begin{array}\
        log(1+z)
        &=int_0^z frac{dt}{1+t}\
        &gt int_0^z frac{dt}{1+z}\
        &=frac{z}{1+z}\
        &=frac{z+z^2-z^2}{1+z}\
        &=z-frac{z^2}{1+z}\
        &gt z-z^2\
        end{array}
        $






        share|cite|improve this answer









        $endgroup$



        All you need is
        $log(1+z) < z$
        for $z > 0$.



        Then
        $log(x+2)-log(x)
        =log(1+frac{2}{x})
        lt frac{2}{x}
        =O(frac1{x})
        $        .



        To show
        $log(1+z) < z$
        for $z > 0$:



        $log(1+z)
        =int_0^z frac{dt}{1+t}
        lt int_0^z dt
        =z
        $        .



        You can also use the integral
        to get a lower bound:



        $begin{array}\
        log(1+z)
        &=int_0^z frac{dt}{1+t}\
        &gt int_0^z frac{dt}{1+z}\
        &=frac{z}{1+z}\
        &=frac{z+z^2-z^2}{1+z}\
        &=z-frac{z^2}{1+z}\
        &gt z-z^2\
        end{array}
        $







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 5:09









        marty cohenmarty cohen

        76.1k549130




        76.1k549130






























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