1 Introduction and summary of results
Write
\(H_{n}\) for the harmonic sum
\(\sum_{r=1}^{n} \frac{1}{r}\). The following well-known estimation can be established by an Euler–Maclaurin summation, or by the logarithmic and binomial series:
$$ H_{n} - \gamma= \log n + \frac{1}{2n} - \frac{1}{12n^{2}} + r_{n}, $$
(1)
where
γ is Euler’s constant and
$$ 0 < r_{n} \leq\frac{1}{120n^{4}}. $$
Since
\(\log(n + \frac{1}{2}) = \log n + \frac{1}{2n} + O( \frac{1}{n^{2}})\), it is a natural idea to absorb the term
\(\frac{1}{2n}\) into the logarithmic term and compare
\(H_{n} - \gamma\) directly with
\(\log(n+ \frac{1}{2})\). This was done by De Temple [
6]: he showed that
$$ H_{n} - \gamma= \log\biggl(n +\frac{1}{2}\biggr) + \frac{1}{24(n+ \frac{1}{2})^{2}} - r_{n}, $$
(2)
where
$$ \frac{7}{960(n+1)^{4}} \leq r_{n} \leq\frac{7}{960n^{4}}. $$
(We will repeatedly re-use the notation
\(r_{n}\) for the remainder term in estimations like this, with a new meaning each time.)
Negoi [
9] demonstrated that the
\(n^{-2}\) term can be absorbed into the log term by considering
\(\log h(n)\), where
\(h(n) = n + \frac{1}{2}+ \frac{1}{24n}\). His result is
$$ H_{n} - \gamma= \log h(n) - r_{n}, $$
(3)
where
\(\frac{1}{48}(n+1)^{-3} \leq r_{n} \leq\frac{1}{48}n^{-3}\). Numerous more recent articles have developed this process further. For example, Chen and Mortici [
3] show that
$$ H_{n} - \gamma= \log \biggl( n + \frac{1}{2}+ \frac{1}{24n} - \frac{1}{48n ^{2}} + \frac{23}{5760n^{3}} \biggr) + O\bigl(n^{-5}\bigr). $$
(4)
Further variations and extensions are given, for example, in [
7] and [
4], and in other references listed in these papers. One natural extension is the replacement of
\(H_{n} - \gamma\) by
\(\psi(x+1)\), where
\(\psi(x)\) is the digamma function
\(\varGamma'(x)/ \varGamma(x)\), since
\(H_{n} - \gamma= \psi(n+1)\).
A slightly different approach, which has proved quite effective, is to compare
\(H_{n} - \gamma\) or
\(\psi(x+1)\) with expressions of the form
\(\frac{1}{2}\log f(n)\). Using no more than the elementary inequalities
\(H_{n} - \gamma\leq\log n + \frac{1}{2n}\) and
\(e^{x} \leq1 + x + x ^{2}\), it is easily shown that
\(H_{n} - \gamma\leq\frac{1}{2}\log(n^{2} + n + 1)\). With
\(h(n)\) as in (
3), we have
\(h(n)^{2} = n^{2} + n + \frac{1}{3} + O(\frac{1}{n})\), suggesting that the right comparison is with
\(f(n) = n^{2} + n + \frac{1}{3}\). Indeed, with
\(f(n)\) defined in this way, Batir [
2] and Lu [
5] have shown, by different methods, that
$$ H_{n} - \gamma= \frac{1}{2} \log f(n) - r_{n}, $$
(5)
where
\(r_{n} \sim1/(180n^{4})\). Lu obtained (
5) as one case of a rather complicated analysis extending to further terms and parameters.
Our objective here is to refine and extend (
5) by considering the sum
$$ H_{n}(x) = \sum_{r=1}^{n} \frac{1}{r+x}, $$
(6)
To identify the limit of
\(H_{n}(x) - \log n\), recall Euler’s limit formula for the Gamma function: this can be written as
\(\varGamma(x+1) = \lim_{n\to\infty} G_{n}(x+1)\), where
$$ G_{n}(x+1) = \frac{n^{x}(n+1)!}{(x+1) \ldots(x+n)}, $$
from which it follows that
\(\lim_{n\to\infty} [H_{n}(x) - \log n] = -\psi(x+1)\). (No further facts about
\(\psi(x)\) are needed for our purposes.) We compare the difference
\(H_{n}(x) + \psi(x+1)\) with
\(\frac{1}{2}\log f(n+x)\). The case
\(x = 0\) reproduces (
5), while the case
\(n = 0\) (with
\(H_{0}(x) =0\)) gives a similar estimate for
\(\psi(x+1)\). Unlike [
2] and [
5], we give explicit upper and lower bounds for the error term, which we will require in the subsequent application. The exact statement is as follows.
Our proof, given in Sect.
2, is a development of De Temple’s original method.
Note that the difference between the upper and lower bounds in (
10) is less than
\(1/45n^{5}\). Of course, (
9) is actually a special case of (
11).
The case
\(x = -\frac{1}{2}\) in (
7) leads to the following estimation of sums of odd reciprocals (which could not be derived from (
9)). The proof is very short, so we include it here.
Theorem
1 has a rather surprising application to an expression that arises in the Dirichlet divisor problem. Denote the divisor function by
\(\tau(n)\) and its summation function
\(\sum_{n \leq x} \tau(n)\) by
\(T(x)\). Write
$$ F(x) = x \log x + (2\gamma-1)x $$
and
$$ T(x) = F(x) + \Delta(x). $$
The most basic form of Dirichlet’s theorem (e.g. [
1, p. 59]) states that
\(\Delta(x) = O(x^{1/2})\). The problem of determining the true order of magnitude of
\(\Delta(x)\) is the “Dirichlet divisor problem”. Denote by
\(\theta_{0}\) the infimum of numbers
θ such that
\(\Delta(x) = O(x^{\theta})\). It was already shown by Voronoi in 1903 that
\(\theta_{0} \leq\frac{1}{3}\) (e.g. see [
10, Sect. 1.6.4]). The estimate has been gradually reduced in a long series of studies: the current best value [
8] is
\(\theta_{0} \leq\frac{131}{416}\).
Write
\([x]\) for the integer part of
x and let
$$ B(x) = x - [x] - \frac{1}{2}, $$
the 1-periodic extension of the function
\(x - \frac{1}{2}\) on
\([0, 1)\). Further, let
$$ S(x) = 2 \sum_{j \leq x^{1/2}} B \biggl( \frac{x}{j} \biggr). $$
Note that
\(|B(x)| \leq\frac{1}{2}\), and hence
\(|S(x)| \leq [x^{1/2}]\), for all
\(x > 0\). A key, if small, step in the proof of Voronoi’s theorem and later refinements is the statement
$$ \Delta(x) = -S(x) + q(x), $$
(13)
where
\(q(x)\) is bounded. A version of the usual proof can be found in [
10, Sect. 1.6.4]; if scrutinised carefully, it gives the bound 3 for
\(|q(x)|\). This is quite good enough for the purpose of proving Voronoi’s theorem: the serious work is the estimation of
\(S(x)\) by exponential sums. However, it is still of some interest to determine the true bounds for
\(q(x)\), along with some other facts about its nature. We will see that
\(q(x)\) is continuous at integers, and that (
9), together with (
1), is exactly what is needed to establish:
3 The remainder in the divisor problem
We return to the divisor problem. The starting point is Dirichlet’s hyperbola identity [
1, p. 59]:
$$ T(x) = 2 \sum_{j \leq x^{1/2}} \biggl[ \frac{x}{j} \biggr] - \bigl[x^{1/2}\bigr]^{2}. $$
(19)
With our previous notation, note that
$$ T(x) + S(x) = F(x) + q(x). $$
Write
\(I_{n} = [n^{2}, (n+1)^{2}]\).
So in fact
\(F(x) + q(x)\) is linear on
\(I_{n}\). For example,
$$ F(x) + q(x) = \textstyle\begin{cases} 2x-2 & \mbox{for } 1 \leq x \leq4, \\ 3x-6 & \mbox{for } 4 \leq x \leq9. \end{cases} $$
The reason for continuity of \(T(x) + S(x)\) is easily seen directly. At non-square integers k, \(T(x)\) increases by \(\tau(k)\). Meanwhile, for each divisor j of k with \(j < k^{1/2}\), \([x/j]\) increases by 1, so \(B(x/j)\) decreases by 1. There are \(\frac{1}{2}\tau(k)\) such divisors j, so \(S(x)\) decreases by \(\tau(k)\). At square integers \(k = n^{2}\), the new term \(2B(k/n) = -1\) enters the sum, so again the decrease in \(S(x)\) is \(\tau(k)\).
To determine the lower bound of
\(q(x)\), we consider
\(q(n^{2})\) and apply (
1).
Finally, we apply (
9) and (
10) to identify the upper bound.
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