5
$\begingroup$

I would like to determine an asymptotic expansion for the following double summation:

$$\sum_{a=1}^{N/\sqrt {j}} \sum_{b=a}^{ja} \frac{1}{ab}$$ where $j$ is a real number $\geq 1$ and $N$ tends to $\infty$. In practice, the summation includes all pairs of integers $a,b$ (with $a \leq b \leq ja $) such that the product $ab$ is $\leq N^2 $.

For $j=1$, the summation yields the infinite harmonic sum of squares, i.e. $\pi^2/6$. For $j>1$, the asymptotic expansion has the form $\log(N) \log(j) + O(1)$, where the $O(1)$ term converges to a value $k$ that depends on $j$. Interestingly, plotting $k$ vs $j$, the resulting function is discontinuous, with the most evident discontinuities occurring for $j$ integer. Is there any way to express this value explicitly? I am not necessarily searching a closed form, which may probably not exist. Rather, I would be interested to know potential alternative ways (e.g., using functions, series, and so on) to express the value of $k$, different from the trivial definition given by the difference between the double summation and the log term.

This question is a general formulation of a problem that is related to this and this other questions previously posted on MSE, for which no conclusive answer was provided despite multiple bounties.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
9
$\begingroup$

Using the decomposition $$ H(x) := \sum_{a \leq x} \frac{1}{a} = \log(x) + \gamma - \frac{\psi(x)}{x} + \int_{x}^{+\infty} \frac{\psi(t) d t}{t^2}, $$ where $\psi(t) = \{ t \} - \frac{1}{2}$, one gets $$ k(j) = - \sum_{n \geq 1} \frac{\psi(nj)}{n^2j } + \tilde{k}(j), $$ where $\tilde{k}$ is an explicit Lipschitz function (see below). In particular :

  • if $j$ is irrational, then $k$ is continuous at $j$.
  • if $j = \frac{a}{q}$ is rational (with $a$ and $q$ coprime), then $k$ has a jump discontinuity, with $k(j^+) = k(j)$, and $k(j^+) - k(j^-) = \frac{\zeta(2)}{aq}$.

Moreover, $\tilde{k}$ has the form $$ \tilde{k}(j)= -\frac{1}{2} \log(j)^2 + \gamma \log(j) + \frac{1}{j} \int_{1}^{+ \infty} \frac{\psi(jt) H(t) d t}{t^2} + c, $$ for some constant $c$.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Thank you for your interesting answer, which provides a nice transformation of the initial summation. Just a question: could the two non-closed elements of your solution be further transformed/simplified? I refer to the summation $\sum_{n \geq 1} \frac{\psi(nj)}{n^2j }$ (or even the simpler $\sum_{n \geq 1} \frac{\{nj\}}{n^2 }$, where $ \{nj\} $ indicates the fractional part), and to the integral included in $\tilde{k}(j)$. May these potential further transformations be facilitated if we set a definite value for $j $, e.g. $j=\sqrt {3} $? $\endgroup$
    – Anatoly
    Commented Sep 15, 2016 at 13:10
  • $\begingroup$ Of course, it can be further transformed, but I doubt it will get significantly "simpler". $\endgroup$
    – js21
    Commented Sep 15, 2016 at 13:17
  • $\begingroup$ Surely. However, regardless of the simplicity, which kind of transformations would you consider in this context? $\endgroup$
    – Anatoly
    Commented Sep 15, 2016 at 13:39
  • $\begingroup$ For example, you can expand $\psi(nj)$ in Fourier series, and get an expression in terms of values of the dilogarithm $Li_2$ on the unit circle, or expand further $H(t)$ in the integral expression of $\tilde{k}$ (this allows a more detailed study of the regularity of $\tilde{k}$),etc. $\endgroup$
    – js21
    Commented Sep 15, 2016 at 15:14
  • $\begingroup$ To give an example of the utility of your calculations, could you please provide the results of your formula for trivial cases, e.g. for $j=1$ or $j=2$? For instance, for $j=1$, the value of $k$ reduces to the harmonic sum of odd squares, giving $\pi^2/6$. Using your formula, the first component of $k (j) $ (the summation) gives $\pi^2/12$. Should then the second component $\tilde k (j) $ be $\pi^2/12$ as well? If so, the value of the integral should also be $\pi^2/12$. $\endgroup$
    – Anatoly
    Commented Sep 26, 2016 at 12:21
2
$\begingroup$

Maple 2016 finds it for concrete values of $j$. For example,

asympt(sum(sum(1/(a*b), b = a .. 3*a, parametric), a = 1 .. N/sqrt(3)), N, 2)

$$\ln \left( 3 \right) \ln \left( N \right) +\ln \left( 3 \right) \ln \left( 1/3\,\sqrt {3} \right) +1/18\,{\pi}^{2}+$$ $$\sum _{a=1}^{ \infty }\,{\frac {\Psi \left( a+2/3 \right) +\Psi \left( a+1/3 \right) -2\,\Psi \left( a \right) }{3a}}+\ln \left( 3 \right) \gamma + {\frac {\,\ln \left( 3 \right) \sqrt {3}-1/3\,\sqrt {3}}{2N}}+O \left( {N}^{-2} \right),$$ where $\Psi(x)$ is the digamma function (http://www.maplesoft.com/support/help/Maple/view.aspx?path=Psi&term=Psi) and $\gamma=0.57721...$ is the Euler-Mascheroni constant.

PS. Also the Mathematica command 

Series[Sum[Sum[1/(a*b), {b, a, 3 a}], {a, 1, N/Sqrt[3]}, Assumptions -> N > 0], {N, Infinity, 1}]

does that in a somewhat more complicated form.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Thanks, this alternative transformation is very interesting. $\endgroup$
    – Anatoly
    Commented Sep 18, 2016 at 17:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .