The sum $sum_{n=1}^infty min_{kle n}{alpha k}$ for irrational $alpha$












8












$begingroup$


Let $alpha$ be an irrational number.
For every $n$ let $z_n$ be the integer closest to the number $alpha n$. Then we can define
$$A(alpha):= sum_{n=1}^infty |alpha n - z_n|.$$



We can also define somewhat similar quantity
$$B(alpha):= sum_{n=1}^infty left(alpha n - lfloor alpha n rfloor right).$$



$A(alpha)$ and $B(alpha)$ are sums of non-negative numbers, so the values can be real numbers of $+infty$.





EDIT: In fact, the above sums are clearly both equal to $+infty$, since $n alpha$ will be infinitely often arbitrarily close to and integer (and also to a number of the form $k+frac12$, $kinmathbb Z$). See, for example, Multiples of an irrational number forming a dense subset (and several other posts on MSE).



I have missed this obvious fact when posting the question. However the sums $A'(alpha)$ and $B'(alpha)$ described below might still be interesting.



I left here the part about $A(alpha)$ and $B(alpha)$, since some users already posted some comments about these sums. If I edited the first part of the post away, those comments would not make sense.





We can also make modifications where we replace the sequence with monotone sequences:
$$
begin{align*}
A'(alpha):=& sum_{n=1}^infty min_{kle n}|alpha k - z_k|;\
B'(alpha):=& sum_{n=1}^infty min_{kle n} left(alpha k - lfloor alpha k rfloorright).\
end{align*}
$$



In some sense, these quantities tell us how far $alpha$ is from rational numbers. (The monotone versions seem more closely related to question whether $alpha$ is rational or irrational, since for rational $alpha$ all but finitely many terms are zero, so the sum must be finite.)



My question is:




  • Were the numbers $A'(alpha)$, $B'(alpha)$ studied somewhere?

  • Can they be calculated for some specific irrational numbers? For example, can we calculate them for $alpha=ln 2$, $alpha=sqrt2$, $alpha=e$, $alpha=pi$ or some other well-know irrational numbers?

  • Do we get $A'(alpha)=infty$ or $B'(alpha)=infty$ for some numbers? If yes, can such numbers be characterized?

  • Are they related to irrationality measure or some other ways to measure how irrational a given number $alpha$ is?




To explain what lead me to these sums: I was thinking about question whether there is a counterexample to Minkowski's theorem for unbounded sets. (Such question was asked here. In the meantime I learned here that such example cannot be found.)
So I was trying to take a line $y=alpha x$ which does not contain any lattice points. I tried to replace the segments of this line by wider rectangles - this lead me to try to compute how wide rectangles I can add without hitting some lattice point. This is related to the question how far are the points $(n,nalpha)$ from lattice points.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You may wish to have a look at Pisot numbers.
    $endgroup$
    – A.P.
    May 2 '15 at 12:41












  • $begingroup$
    @MichaelBurr I meant to evaluate them. (I.e., if they are $+infty$ for each $alpha$, that would answer my question.) However, I do not think that $A'(alpha)$ and $B'(alpha)$ are always infinite.
    $endgroup$
    – Martin Sleziak
    May 2 '15 at 12:44






  • 1




    $begingroup$
    $A$ and $B$ should be infinite for all nonintegral $alpha$ (they are periodic for $mathbb{Q}setminusmathbb{Z}$). The other two sums are more interesting (indeed)
    $endgroup$
    – Michael Burr
    May 2 '15 at 12:46








  • 1




    $begingroup$
    It is not obvious to me that $A(alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $alpha$ then $|alpha n-z_n|$ should behave like $frac{1}{n}$, leading to a divergent sum.
    $endgroup$
    – Jack D'Aurizio
    May 2 '15 at 13:02






  • 2




    $begingroup$
    I bet you can compute these sums for simple quadratic irrationals such as the golden ratio.
    $endgroup$
    – Gerry Myerson
    May 9 '15 at 9:15
















8












$begingroup$


Let $alpha$ be an irrational number.
For every $n$ let $z_n$ be the integer closest to the number $alpha n$. Then we can define
$$A(alpha):= sum_{n=1}^infty |alpha n - z_n|.$$



We can also define somewhat similar quantity
$$B(alpha):= sum_{n=1}^infty left(alpha n - lfloor alpha n rfloor right).$$



$A(alpha)$ and $B(alpha)$ are sums of non-negative numbers, so the values can be real numbers of $+infty$.





EDIT: In fact, the above sums are clearly both equal to $+infty$, since $n alpha$ will be infinitely often arbitrarily close to and integer (and also to a number of the form $k+frac12$, $kinmathbb Z$). See, for example, Multiples of an irrational number forming a dense subset (and several other posts on MSE).



I have missed this obvious fact when posting the question. However the sums $A'(alpha)$ and $B'(alpha)$ described below might still be interesting.



I left here the part about $A(alpha)$ and $B(alpha)$, since some users already posted some comments about these sums. If I edited the first part of the post away, those comments would not make sense.





We can also make modifications where we replace the sequence with monotone sequences:
$$
begin{align*}
A'(alpha):=& sum_{n=1}^infty min_{kle n}|alpha k - z_k|;\
B'(alpha):=& sum_{n=1}^infty min_{kle n} left(alpha k - lfloor alpha k rfloorright).\
end{align*}
$$



In some sense, these quantities tell us how far $alpha$ is from rational numbers. (The monotone versions seem more closely related to question whether $alpha$ is rational or irrational, since for rational $alpha$ all but finitely many terms are zero, so the sum must be finite.)



My question is:




  • Were the numbers $A'(alpha)$, $B'(alpha)$ studied somewhere?

  • Can they be calculated for some specific irrational numbers? For example, can we calculate them for $alpha=ln 2$, $alpha=sqrt2$, $alpha=e$, $alpha=pi$ or some other well-know irrational numbers?

  • Do we get $A'(alpha)=infty$ or $B'(alpha)=infty$ for some numbers? If yes, can such numbers be characterized?

  • Are they related to irrationality measure or some other ways to measure how irrational a given number $alpha$ is?




To explain what lead me to these sums: I was thinking about question whether there is a counterexample to Minkowski's theorem for unbounded sets. (Such question was asked here. In the meantime I learned here that such example cannot be found.)
So I was trying to take a line $y=alpha x$ which does not contain any lattice points. I tried to replace the segments of this line by wider rectangles - this lead me to try to compute how wide rectangles I can add without hitting some lattice point. This is related to the question how far are the points $(n,nalpha)$ from lattice points.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You may wish to have a look at Pisot numbers.
    $endgroup$
    – A.P.
    May 2 '15 at 12:41












  • $begingroup$
    @MichaelBurr I meant to evaluate them. (I.e., if they are $+infty$ for each $alpha$, that would answer my question.) However, I do not think that $A'(alpha)$ and $B'(alpha)$ are always infinite.
    $endgroup$
    – Martin Sleziak
    May 2 '15 at 12:44






  • 1




    $begingroup$
    $A$ and $B$ should be infinite for all nonintegral $alpha$ (they are periodic for $mathbb{Q}setminusmathbb{Z}$). The other two sums are more interesting (indeed)
    $endgroup$
    – Michael Burr
    May 2 '15 at 12:46








  • 1




    $begingroup$
    It is not obvious to me that $A(alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $alpha$ then $|alpha n-z_n|$ should behave like $frac{1}{n}$, leading to a divergent sum.
    $endgroup$
    – Jack D'Aurizio
    May 2 '15 at 13:02






  • 2




    $begingroup$
    I bet you can compute these sums for simple quadratic irrationals such as the golden ratio.
    $endgroup$
    – Gerry Myerson
    May 9 '15 at 9:15














8












8








8


7



$begingroup$


Let $alpha$ be an irrational number.
For every $n$ let $z_n$ be the integer closest to the number $alpha n$. Then we can define
$$A(alpha):= sum_{n=1}^infty |alpha n - z_n|.$$



We can also define somewhat similar quantity
$$B(alpha):= sum_{n=1}^infty left(alpha n - lfloor alpha n rfloor right).$$



$A(alpha)$ and $B(alpha)$ are sums of non-negative numbers, so the values can be real numbers of $+infty$.





EDIT: In fact, the above sums are clearly both equal to $+infty$, since $n alpha$ will be infinitely often arbitrarily close to and integer (and also to a number of the form $k+frac12$, $kinmathbb Z$). See, for example, Multiples of an irrational number forming a dense subset (and several other posts on MSE).



I have missed this obvious fact when posting the question. However the sums $A'(alpha)$ and $B'(alpha)$ described below might still be interesting.



I left here the part about $A(alpha)$ and $B(alpha)$, since some users already posted some comments about these sums. If I edited the first part of the post away, those comments would not make sense.





We can also make modifications where we replace the sequence with monotone sequences:
$$
begin{align*}
A'(alpha):=& sum_{n=1}^infty min_{kle n}|alpha k - z_k|;\
B'(alpha):=& sum_{n=1}^infty min_{kle n} left(alpha k - lfloor alpha k rfloorright).\
end{align*}
$$



In some sense, these quantities tell us how far $alpha$ is from rational numbers. (The monotone versions seem more closely related to question whether $alpha$ is rational or irrational, since for rational $alpha$ all but finitely many terms are zero, so the sum must be finite.)



My question is:




  • Were the numbers $A'(alpha)$, $B'(alpha)$ studied somewhere?

  • Can they be calculated for some specific irrational numbers? For example, can we calculate them for $alpha=ln 2$, $alpha=sqrt2$, $alpha=e$, $alpha=pi$ or some other well-know irrational numbers?

  • Do we get $A'(alpha)=infty$ or $B'(alpha)=infty$ for some numbers? If yes, can such numbers be characterized?

  • Are they related to irrationality measure or some other ways to measure how irrational a given number $alpha$ is?




To explain what lead me to these sums: I was thinking about question whether there is a counterexample to Minkowski's theorem for unbounded sets. (Such question was asked here. In the meantime I learned here that such example cannot be found.)
So I was trying to take a line $y=alpha x$ which does not contain any lattice points. I tried to replace the segments of this line by wider rectangles - this lead me to try to compute how wide rectangles I can add without hitting some lattice point. This is related to the question how far are the points $(n,nalpha)$ from lattice points.










share|cite|improve this question











$endgroup$




Let $alpha$ be an irrational number.
For every $n$ let $z_n$ be the integer closest to the number $alpha n$. Then we can define
$$A(alpha):= sum_{n=1}^infty |alpha n - z_n|.$$



We can also define somewhat similar quantity
$$B(alpha):= sum_{n=1}^infty left(alpha n - lfloor alpha n rfloor right).$$



$A(alpha)$ and $B(alpha)$ are sums of non-negative numbers, so the values can be real numbers of $+infty$.





EDIT: In fact, the above sums are clearly both equal to $+infty$, since $n alpha$ will be infinitely often arbitrarily close to and integer (and also to a number of the form $k+frac12$, $kinmathbb Z$). See, for example, Multiples of an irrational number forming a dense subset (and several other posts on MSE).



I have missed this obvious fact when posting the question. However the sums $A'(alpha)$ and $B'(alpha)$ described below might still be interesting.



I left here the part about $A(alpha)$ and $B(alpha)$, since some users already posted some comments about these sums. If I edited the first part of the post away, those comments would not make sense.





We can also make modifications where we replace the sequence with monotone sequences:
$$
begin{align*}
A'(alpha):=& sum_{n=1}^infty min_{kle n}|alpha k - z_k|;\
B'(alpha):=& sum_{n=1}^infty min_{kle n} left(alpha k - lfloor alpha k rfloorright).\
end{align*}
$$



In some sense, these quantities tell us how far $alpha$ is from rational numbers. (The monotone versions seem more closely related to question whether $alpha$ is rational or irrational, since for rational $alpha$ all but finitely many terms are zero, so the sum must be finite.)



My question is:




  • Were the numbers $A'(alpha)$, $B'(alpha)$ studied somewhere?

  • Can they be calculated for some specific irrational numbers? For example, can we calculate them for $alpha=ln 2$, $alpha=sqrt2$, $alpha=e$, $alpha=pi$ or some other well-know irrational numbers?

  • Do we get $A'(alpha)=infty$ or $B'(alpha)=infty$ for some numbers? If yes, can such numbers be characterized?

  • Are they related to irrationality measure or some other ways to measure how irrational a given number $alpha$ is?




To explain what lead me to these sums: I was thinking about question whether there is a counterexample to Minkowski's theorem for unbounded sets. (Such question was asked here. In the meantime I learned here that such example cannot be found.)
So I was trying to take a line $y=alpha x$ which does not contain any lattice points. I tried to replace the segments of this line by wider rectangles - this lead me to try to compute how wide rectangles I can add without hitting some lattice point. This is related to the question how far are the points $(n,nalpha)$ from lattice points.







sequences-and-series irrational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:58









Community

1




1










asked May 2 '15 at 12:31









Martin SleziakMartin Sleziak

44.7k9117272




44.7k9117272








  • 1




    $begingroup$
    You may wish to have a look at Pisot numbers.
    $endgroup$
    – A.P.
    May 2 '15 at 12:41












  • $begingroup$
    @MichaelBurr I meant to evaluate them. (I.e., if they are $+infty$ for each $alpha$, that would answer my question.) However, I do not think that $A'(alpha)$ and $B'(alpha)$ are always infinite.
    $endgroup$
    – Martin Sleziak
    May 2 '15 at 12:44






  • 1




    $begingroup$
    $A$ and $B$ should be infinite for all nonintegral $alpha$ (they are periodic for $mathbb{Q}setminusmathbb{Z}$). The other two sums are more interesting (indeed)
    $endgroup$
    – Michael Burr
    May 2 '15 at 12:46








  • 1




    $begingroup$
    It is not obvious to me that $A(alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $alpha$ then $|alpha n-z_n|$ should behave like $frac{1}{n}$, leading to a divergent sum.
    $endgroup$
    – Jack D'Aurizio
    May 2 '15 at 13:02






  • 2




    $begingroup$
    I bet you can compute these sums for simple quadratic irrationals such as the golden ratio.
    $endgroup$
    – Gerry Myerson
    May 9 '15 at 9:15














  • 1




    $begingroup$
    You may wish to have a look at Pisot numbers.
    $endgroup$
    – A.P.
    May 2 '15 at 12:41












  • $begingroup$
    @MichaelBurr I meant to evaluate them. (I.e., if they are $+infty$ for each $alpha$, that would answer my question.) However, I do not think that $A'(alpha)$ and $B'(alpha)$ are always infinite.
    $endgroup$
    – Martin Sleziak
    May 2 '15 at 12:44






  • 1




    $begingroup$
    $A$ and $B$ should be infinite for all nonintegral $alpha$ (they are periodic for $mathbb{Q}setminusmathbb{Z}$). The other two sums are more interesting (indeed)
    $endgroup$
    – Michael Burr
    May 2 '15 at 12:46








  • 1




    $begingroup$
    It is not obvious to me that $A(alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $alpha$ then $|alpha n-z_n|$ should behave like $frac{1}{n}$, leading to a divergent sum.
    $endgroup$
    – Jack D'Aurizio
    May 2 '15 at 13:02






  • 2




    $begingroup$
    I bet you can compute these sums for simple quadratic irrationals such as the golden ratio.
    $endgroup$
    – Gerry Myerson
    May 9 '15 at 9:15








1




1




$begingroup$
You may wish to have a look at Pisot numbers.
$endgroup$
– A.P.
May 2 '15 at 12:41






$begingroup$
You may wish to have a look at Pisot numbers.
$endgroup$
– A.P.
May 2 '15 at 12:41














$begingroup$
@MichaelBurr I meant to evaluate them. (I.e., if they are $+infty$ for each $alpha$, that would answer my question.) However, I do not think that $A'(alpha)$ and $B'(alpha)$ are always infinite.
$endgroup$
– Martin Sleziak
May 2 '15 at 12:44




$begingroup$
@MichaelBurr I meant to evaluate them. (I.e., if they are $+infty$ for each $alpha$, that would answer my question.) However, I do not think that $A'(alpha)$ and $B'(alpha)$ are always infinite.
$endgroup$
– Martin Sleziak
May 2 '15 at 12:44




1




1




$begingroup$
$A$ and $B$ should be infinite for all nonintegral $alpha$ (they are periodic for $mathbb{Q}setminusmathbb{Z}$). The other two sums are more interesting (indeed)
$endgroup$
– Michael Burr
May 2 '15 at 12:46






$begingroup$
$A$ and $B$ should be infinite for all nonintegral $alpha$ (they are periodic for $mathbb{Q}setminusmathbb{Z}$). The other two sums are more interesting (indeed)
$endgroup$
– Michael Burr
May 2 '15 at 12:46






1




1




$begingroup$
It is not obvious to me that $A(alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $alpha$ then $|alpha n-z_n|$ should behave like $frac{1}{n}$, leading to a divergent sum.
$endgroup$
– Jack D'Aurizio
May 2 '15 at 13:02




$begingroup$
It is not obvious to me that $A(alpha)$ is well-defined. Even if $n$ is the denominator of a convergent of the continued fraction of $alpha$ then $|alpha n-z_n|$ should behave like $frac{1}{n}$, leading to a divergent sum.
$endgroup$
– Jack D'Aurizio
May 2 '15 at 13:02




2




2




$begingroup$
I bet you can compute these sums for simple quadratic irrationals such as the golden ratio.
$endgroup$
– Gerry Myerson
May 9 '15 at 9:15




$begingroup$
I bet you can compute these sums for simple quadratic irrationals such as the golden ratio.
$endgroup$
– Gerry Myerson
May 9 '15 at 9:15










1 Answer
1






active

oldest

votes


















0












$begingroup$

$newcommand{bivec}[2]{begin{bmatrix} #1 \ #2 end{bmatrix}}$
Excavating. Still the series for $A'(alpha)$ and $B'(alpha)$ both diverge always.



Let $alpha=[a_0;a_1,a_2,ldots]$ be the continued fraction expansion of $alpha$, and $p_n/q_n$ be its $n$-th convergent:
$$bivec{p_{-1}}{q_{-1}}=bivec{1}{0}, bivec{p_0}{q_0}=bivec{a_0}{1}, bivec{p_n}{q_n}=a_nbivec{p_{n-1}}{q_{n-1}}+bivec{p_{n-2}}{q_{n-2}}.$$
Then for $rgeqslant 0$ and $q_rleqslant n<q_{r+1}$ we have $displaystylemin_{kleqslant n}|alpha k-z_k|=|alpha q_r-p_r|$, which gives
$$A'(alpha)=sum_{n=0}^{infty}(q_{n+1}-q_n)cdot|alpha q_n-p_n|>sum_{n=0}^{infty}frac{q_{n+1}-q_n}{q_{n+1}+q_n}$$
by "Theorem 5" here. But $r_n=q_{n+1}/q_n$ can't converge to $1$ because $r_n=a_{n+1}+1/r_{n-1}$.



(Therefore $dfrac{q_{n+1}-q_n}{q_{n+1}+q_n}=dfrac{r_n-1}{r_n+1}$ can't converge to $0$.)



Thus $A'(alpha)=+infty$ for any (irrational) $alpha$, and therefore $B'(alpha)=+infty$ too.






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    active

    oldest

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    0












    $begingroup$

    $newcommand{bivec}[2]{begin{bmatrix} #1 \ #2 end{bmatrix}}$
    Excavating. Still the series for $A'(alpha)$ and $B'(alpha)$ both diverge always.



    Let $alpha=[a_0;a_1,a_2,ldots]$ be the continued fraction expansion of $alpha$, and $p_n/q_n$ be its $n$-th convergent:
    $$bivec{p_{-1}}{q_{-1}}=bivec{1}{0}, bivec{p_0}{q_0}=bivec{a_0}{1}, bivec{p_n}{q_n}=a_nbivec{p_{n-1}}{q_{n-1}}+bivec{p_{n-2}}{q_{n-2}}.$$
    Then for $rgeqslant 0$ and $q_rleqslant n<q_{r+1}$ we have $displaystylemin_{kleqslant n}|alpha k-z_k|=|alpha q_r-p_r|$, which gives
    $$A'(alpha)=sum_{n=0}^{infty}(q_{n+1}-q_n)cdot|alpha q_n-p_n|>sum_{n=0}^{infty}frac{q_{n+1}-q_n}{q_{n+1}+q_n}$$
    by "Theorem 5" here. But $r_n=q_{n+1}/q_n$ can't converge to $1$ because $r_n=a_{n+1}+1/r_{n-1}$.



    (Therefore $dfrac{q_{n+1}-q_n}{q_{n+1}+q_n}=dfrac{r_n-1}{r_n+1}$ can't converge to $0$.)



    Thus $A'(alpha)=+infty$ for any (irrational) $alpha$, and therefore $B'(alpha)=+infty$ too.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $newcommand{bivec}[2]{begin{bmatrix} #1 \ #2 end{bmatrix}}$
      Excavating. Still the series for $A'(alpha)$ and $B'(alpha)$ both diverge always.



      Let $alpha=[a_0;a_1,a_2,ldots]$ be the continued fraction expansion of $alpha$, and $p_n/q_n$ be its $n$-th convergent:
      $$bivec{p_{-1}}{q_{-1}}=bivec{1}{0}, bivec{p_0}{q_0}=bivec{a_0}{1}, bivec{p_n}{q_n}=a_nbivec{p_{n-1}}{q_{n-1}}+bivec{p_{n-2}}{q_{n-2}}.$$
      Then for $rgeqslant 0$ and $q_rleqslant n<q_{r+1}$ we have $displaystylemin_{kleqslant n}|alpha k-z_k|=|alpha q_r-p_r|$, which gives
      $$A'(alpha)=sum_{n=0}^{infty}(q_{n+1}-q_n)cdot|alpha q_n-p_n|>sum_{n=0}^{infty}frac{q_{n+1}-q_n}{q_{n+1}+q_n}$$
      by "Theorem 5" here. But $r_n=q_{n+1}/q_n$ can't converge to $1$ because $r_n=a_{n+1}+1/r_{n-1}$.



      (Therefore $dfrac{q_{n+1}-q_n}{q_{n+1}+q_n}=dfrac{r_n-1}{r_n+1}$ can't converge to $0$.)



      Thus $A'(alpha)=+infty$ for any (irrational) $alpha$, and therefore $B'(alpha)=+infty$ too.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $newcommand{bivec}[2]{begin{bmatrix} #1 \ #2 end{bmatrix}}$
        Excavating. Still the series for $A'(alpha)$ and $B'(alpha)$ both diverge always.



        Let $alpha=[a_0;a_1,a_2,ldots]$ be the continued fraction expansion of $alpha$, and $p_n/q_n$ be its $n$-th convergent:
        $$bivec{p_{-1}}{q_{-1}}=bivec{1}{0}, bivec{p_0}{q_0}=bivec{a_0}{1}, bivec{p_n}{q_n}=a_nbivec{p_{n-1}}{q_{n-1}}+bivec{p_{n-2}}{q_{n-2}}.$$
        Then for $rgeqslant 0$ and $q_rleqslant n<q_{r+1}$ we have $displaystylemin_{kleqslant n}|alpha k-z_k|=|alpha q_r-p_r|$, which gives
        $$A'(alpha)=sum_{n=0}^{infty}(q_{n+1}-q_n)cdot|alpha q_n-p_n|>sum_{n=0}^{infty}frac{q_{n+1}-q_n}{q_{n+1}+q_n}$$
        by "Theorem 5" here. But $r_n=q_{n+1}/q_n$ can't converge to $1$ because $r_n=a_{n+1}+1/r_{n-1}$.



        (Therefore $dfrac{q_{n+1}-q_n}{q_{n+1}+q_n}=dfrac{r_n-1}{r_n+1}$ can't converge to $0$.)



        Thus $A'(alpha)=+infty$ for any (irrational) $alpha$, and therefore $B'(alpha)=+infty$ too.






        share|cite|improve this answer











        $endgroup$



        $newcommand{bivec}[2]{begin{bmatrix} #1 \ #2 end{bmatrix}}$
        Excavating. Still the series for $A'(alpha)$ and $B'(alpha)$ both diverge always.



        Let $alpha=[a_0;a_1,a_2,ldots]$ be the continued fraction expansion of $alpha$, and $p_n/q_n$ be its $n$-th convergent:
        $$bivec{p_{-1}}{q_{-1}}=bivec{1}{0}, bivec{p_0}{q_0}=bivec{a_0}{1}, bivec{p_n}{q_n}=a_nbivec{p_{n-1}}{q_{n-1}}+bivec{p_{n-2}}{q_{n-2}}.$$
        Then for $rgeqslant 0$ and $q_rleqslant n<q_{r+1}$ we have $displaystylemin_{kleqslant n}|alpha k-z_k|=|alpha q_r-p_r|$, which gives
        $$A'(alpha)=sum_{n=0}^{infty}(q_{n+1}-q_n)cdot|alpha q_n-p_n|>sum_{n=0}^{infty}frac{q_{n+1}-q_n}{q_{n+1}+q_n}$$
        by "Theorem 5" here. But $r_n=q_{n+1}/q_n$ can't converge to $1$ because $r_n=a_{n+1}+1/r_{n-1}$.



        (Therefore $dfrac{q_{n+1}-q_n}{q_{n+1}+q_n}=dfrac{r_n-1}{r_n+1}$ can't converge to $0$.)



        Thus $A'(alpha)=+infty$ for any (irrational) $alpha$, and therefore $B'(alpha)=+infty$ too.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 17:48

























        answered Jan 3 at 11:49









        metamorphymetamorphy

        3,6821621




        3,6821621






























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