Skip to main content

A logarithmic estimate for harmonic sums and the digamma function, with an application to the Dirichlet divisor problem

Abstract

Let \(H_{n} = \sum_{r=1}^{n} 1/r\) and \(H_{n}(x) = \sum_{r=1}^{n} 1/(r+x)\). Let \(\psi(x)\) denote the digamma function. It is shown that \(H_{n}(x) + \psi(x+1)\) is approximated by \(\frac{1}{2}\log f(n+x)\), where \(f(x) = x^{2} + x + \frac{1}{3}\), with error term of order \((n+x)^{-5}\). The cases \(x = 0\) and \(n = 0\) equate to estimates for \(H_{n} - \gamma \) and \(\psi(x+1)\) itself. The result is applied to determine exact bounds for a remainder term occurring in the Dirichlet divisor problem.

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.

Theorem 1

Let \(n \geq1\) and \(x > -1\), or \(n = 0\) and \(x > 0\), and let \(H_{n}(x)\) be defined by (6). Let \(f(x) = x^{2} + x + \frac{1}{3}\). Then

$$ H_{n}(x) + \psi(x+1) = \frac{1}{2} \log f(n+x) - r(n,x), $$
(7)

where

$$ \frac{1}{180(n+x+1)^{4}} \leq r(n,x) \leq\frac{1}{180 (n+x)^{4}}. $$
(8)

In particular,

$$ H_{n} - \gamma= \frac{1}{2}\log f(n) - r_{n}, $$
(9)

where

$$ \frac{1}{180(n+1)^{4}} \leq r_{n} \leq\frac{1}{180 n^{4}}. $$
(10)

Also, for \(x > 0\),

$$ \psi(x+1) = \frac{1}{2}\log f(x) - r(x), $$
(11)

where

$$ \frac{1}{180(x+1)^{4}} \leq r(x) \leq\frac{1}{180 x^{4}}. $$
(12)

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.

Corollary 1

Let \(U_{n} = \sum_{r=1}^{n} \frac{1}{2r-1}\). Then

$$ U_{n} - \frac{1}{2} \gamma- \log2 = \frac{1}{4} \log \biggl(n^{2} + \frac{1}{12}\biggr) - r_{n}, $$

where

$$ \frac{1}{360(n+ \frac{1}{2})^{4}} \leq r_{n} \leq \frac{1}{360(n- \frac{1}{2})^{4}}. $$

Proof

Note that \(2U_{n} = H_{n}(-\frac{1}{2})\). Also, \(2U_{n} = 2H_{2n} - H_{n}\), from which it follows easily that \(-\psi(\frac{1}{2}) = \lim_{n\to \infty }[2U_{n} - \log n] = \gamma+ 2\log2\). Finally, \(f(n- \frac{1}{2}) = n ^{2} + \frac{1}{12}\). □

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:

Theorem 2

With \(q(x)\) defined as above, we have, for all \(x \geq1\),

$$ - \frac{1}{6} < q(x) < \frac{1}{3} . $$
(14)

Both bounds are optimal.

The proof of Theorem 1

The key step is the following lemma.

Lemma 1

Let \(f(x) = x^{2} + x + \frac{1}{3}\). Then, for all \(x \geq1\),

$$ \log f(x) - \log f(x-1) = \frac{2}{x} - \delta(x), $$
(15)

where

$$ \frac{2}{45x^{5}} < \delta(x) < \frac{2}{45(x - \frac{1}{2})^{5}}. $$
(16)

Proof

We start with \(f(x) = x^{2} + x + c\) and allow the choice of c to emerge from the reasoning. Let \(\delta(x)\) be defined by (15). Now \(f(x)/f(x-1) \to1\), and hence \(\delta(x) \to0\), as \(x \to\infty\). So \(\delta(x) = - \int_{x}^{\infty}\delta'(t) \,dt\) for all \(x > 0\). Now

$$\begin{aligned} - \delta'(t) & = \frac{2t+1}{f(t)} - \frac{2t-1}{f(t-1)} + \frac{2}{t ^{2}} \\ & = \frac{G(t)}{t^{2} f(t) f(t-1)}, \end{aligned}$$

where

$$\begin{aligned} G(t) & = t^{2}(2t+1) \bigl(t^{2} - t +c\bigr) - t^{2}(2t-1) \bigl(t^{2} + t + c\bigr) + 2\bigl(t ^{2} + t + c\bigr) \bigl(t^{2} - t + c\bigr) \\ & = -t^{2} \bigl(2t^{2} - 2c\bigr) + 2\bigl[ t^{4} + (2c-1)t^{2} + c^{2}\bigr] \\ & = 2(3c - 1)t^{2} + 2c^{2}. \end{aligned}$$

To eliminate the \(t^{2}\) term, we now choose \(c = \frac{1}{3}\), so that \(G(t) = \frac{2}{9} \) and

$$ -\delta'(t) = \frac{2}{9 t^{2} f(t) f(t-1)}. $$

Now \(f(t) f(t-1) = t^{4} - \frac{1}{3} t^{2} + \frac{1}{9} < t^{4}\) for \(t \geq1\), so

$$ \delta(x) > \int_{x} ^{\infty}\frac{2}{9t^{6}} \,dt = \frac{2}{45x ^{5}} $$

for \(x \geq1\). On the other hand, \(f(t) > f(t-1) > (t - \frac{1}{2})^{2}\), hence

$$ \delta(x) < \int_{x}^{\infty}\frac{2}{9(t- \frac{1}{2})^{6}} \,dt = \frac{2}{45(x - \frac{1}{2})^{5}}. $$

 □

Proof of Theorem 1

Apply the identity (15) to \(r+x\) for \(1 \leq r \leq n\) and add: we find

$$ \log f(n+x) - \log f(x) = 2H_{n}(x) - \sum _{r=1}^{n} \delta(r+x), $$

equivalently

$$ 2H_{n}(x) - \log f(n+x) = -\log f(x) + \sum _{r=1}^{n} \delta(r+x). $$
(17)

Now \(f(n+x)/n^{2} \to1\), so \(\log f(n+x) - 2 \log n \to0\), as \(n \to\infty\). Taking the limit in (17), we see that

$$ -2\psi(x+1) = -\log f(x) + \sum_{r=1}^{\infty} \delta(r+x). $$
(18)

Now taking the difference, we have

$$ 2H_{n}(x) - \log f(n+x) + 2\psi(x+1) = -2r(n,x), $$

where

$$ 2r(n,x) = \sum_{r = n+1}^{\infty}\delta(r+x). $$

The condition \(x > -1\) ensures that the inequality (16) applies to \(\delta(r+x)\) for \(r \geq2\). By integral estimation, we now have, for \(n \geq1\),

$$ r(n,x) \geq\sum_{r = n+1}^{\infty} \frac{1}{45(r+x)^{5}} > \int_{n+1} ^{\infty}\frac{1}{45(t+x)^{5}} \,dt = \frac{1}{180(n+x+1)^{4}}. $$

At the same time,

$$ r(n,x) \leq\sum_{r = n+1}^{\infty} \frac{1}{45(r- \frac{1}{2}+x)^{5}}. $$

The function \(1/(t+x)^{5}\) is convex, and convex functions \(h(t)\) satisfy \(h(y -\frac{1}{2}) \leq\int_{y-1}^{y} h(t) \,dt\), hence

$$ r(n,x) \leq \int_{n}^{\infty}\frac{1}{45(t+x)^{5}} \,dt = \frac{1}{180(n+x)^{4}}. $$

For \(x > 0\), the case \(n = 0\) follows similarly from (18) (note that (16) now applies also to \(\delta(1+x)\)). □

Note 1

The upper bounds for \(\delta(x)\) in Lemma 1 and \(r(n,x)\) in Theorem 1 can be slightly improved. One can verify that \(t^{2} f(t) f(t-1) \geq(t - \frac{1}{12})^{6}\) where we previously used \((t - \frac{1}{2})^{6}\). This leads to \(\delta(x) < 2/[45(x - \frac{1}{12})^{5}]\) and \(r(n,x) \leq1/[180 (n+x+ \frac{5}{12})^{4}]\).

Note 2

In principle, one could derive Stirling-type approximations to \(\log\varGamma(x)\) and \(\varGamma(x)\) from (11), but only in terms of the rather unpleasant antiderivative of \(\log f(x)\).

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 \(I_{n} = [n^{2}, (n+1)^{2}]\).

Lemma 2

For \(x \in I_{n}\),

$$ q(x) = xH_{n} - F(x) - n(n+1). $$
(20)

The function \(q(x)\) is continuous for all \(x \geq1\), and concave on each interval \(I_{n}\).

Proof

We work with \(T(x) + S(x)\). For \(n^{2} \leq x < (n+1)^{2}\), we have by (19)

$$\begin{aligned} T(x) & = 2 \sum_{j=1}^{n} \biggl( \frac{x}{j} - B \biggl( \frac{x}{j} \biggr) - \frac{1}{2} \biggr) - n^{2} \\ & = 2xH_{n} - S(x) - n - n^{2}, \end{aligned}$$

which equates to (20). We check that this remains valid at \(x = (n+1)^{2}\). By what we have just shown, with n replaced by \(n+1\),

$$\begin{aligned} T\bigl[(n+1)^{2}\bigr] + S\bigl[(n+1)^{2}\bigr] & = 2H_{n+1} (n+1)^{2} - (n+1) (n+2) \\ & = 2H_{n} (n+1)^{2} + 2(n+1) - (n+1) (n+2) \\ & = 2H_{n} (n+1)^{2} - n(n+1), \end{aligned}$$

agreeing with (20). Hence \(F(x) + q(x)\), and consequently \(q(x)\) itself, is continuous for all \(x \geq1\). Also, \(q'(x) = 2H_{n} - F'(x) = 2H _{n} - \log x - 2\gamma\), which is decreasing, so \(q(x)\) is concave on \(I_{n}\). □

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).

Lemma 3

We have \(q(x) > - \frac{1}{6} \) for all \(x \geq1\). Further, \(q(n^{2}) \leq-\frac{1}{6} + \frac{1}{60n^{2}}\), so \(\inf_{x\geq1} q(x)= -\frac{1}{6}\).

Proof

By (20) and (1) (with \(r_{n}\) as in (1)),

$$\begin{aligned} q\bigl(n^{2}\bigr) & = 2n^{2} H_{n} - 2n^{2} \log n - (2\gamma-1)n^{2} - n - n^{2} \\ & = 2n^{2} (H_{n} - \log n - \gamma) - n \\ & = 2n^{2} \biggl( \frac{1}{2n} - \frac{1}{12n^{2}} + r_{n} \biggr) - n \\ & = - \frac{1}{6} + 2n^{2} r_{n}. \end{aligned}$$

So

$$ -\frac{1}{6} < q\bigl(n^{2}\bigr) \leq- \frac{1}{6} + \frac{1}{60n^{2}}. $$

This applies also at \((n+1)^{2}\). Since \(q(x)\) is concave on \(I_{n}\), it follows that \(q(x) > - \frac{1}{6} \) throughout this interval. □

Finally, we apply (9) and (10) to identify the upper bound.

Proof of the upper bound in Theorem 2

Let \(x_{n}\) be the point where \(q(x)\) attains its maximum in \(I_{n}\). Since \(q'(x) = 2H_{n} - \log x - 2\gamma\), we have \(\log x_{n} = 2H_{n} - 2\gamma\), hence the maximum value is

$$\begin{aligned} q(x_{n}) & = 2H_{n} x_{n} - F(x_{n}) - n(n+1) \\ & = (\log x_{n} + 2\gamma)x_{n} - F(x_{n}) - n(n+1) \\ & = x_{n} - n(n+1). \end{aligned}$$
(21)

By (9) and (10) (using only \(r_{n} > 0\)), we have \(\log x_{n} < \log f(n)\), so \(x_{n} < f(n) = n(n+1) + \frac{1}{3} \), hence \(q(x_{n}) < \frac{1}{3} \).

We show that, conversely,

$$ q(x_{n}) > \frac{1}{3} - \frac{1}{45n^{2}} $$

for \(n \geq2\), so that \(\sup_{x \geq1} q(x) = \frac{1}{3} \). By (21), this is equivalent to \(x_{n} > f(n) - \frac{1}{45n^{2}}\). By (10),

$$ \log x_{n} = 2H_{n} - 2\gamma\geq\log f(n) - \frac{1}{90n^{4}}. $$

Now using the inequality \(e^{-x} \geq1 - x\), together with \(f(n) \leq2n^{2}\), we have

$$ x_{n} \geq f(n) e^{-1/90n^{4}} \geq f(n) \biggl( 1 - \frac{1}{90n^{4}} \biggr) \geq f(n) - \frac{1}{45n^{2}}. $$

 □

References

  1. Apostol, T.M.: Introduction to Analytic Number Theory. Springer, Berlin (1976)

    MATH  Google Scholar 

  2. Batir, N.: Sharp bounds for the psi function and harmonic numbers. Math. Inequal. Appl. 14, 917–925 (2011)

    MathSciNet  MATH  Google Scholar 

  3. Chen, C., Mortici, C.: New sequence converging towards the Euler–Mascheroni constant. Comput. Math. Appl. 64, 391–398 (2012)

    MathSciNet  Article  Google Scholar 

  4. Dawei, L.: Some new convergent sequences and inequalities of Euler’s constant. J. Math. Anal. Appl. 419, 541–552 (2014)

    MathSciNet  Article  Google Scholar 

  5. Dawei, L.: Some quicker classes of sequences convergent to Euler’s constant. Appl. Math. Comput. 232, 172–177 (2014)

    MathSciNet  MATH  Google Scholar 

  6. De Temple, D.W.: A quicker convergence to Euler’s constant. Am. Math. Mon. 100, 468–470 (1993)

    MathSciNet  Article  Google Scholar 

  7. Elezovic, N.: Estimations of psi function and harmonic numbers. Appl. Math. Comput. 258, 192–205 (2015)

    MathSciNet  MATH  Google Scholar 

  8. Huxley, M.N.: Exponential sums and lattice points III. Proc. Lond. Math. Soc. (3) 87, 591–609 (2003)

    MathSciNet  Article  Google Scholar 

  9. Negoi, T.: A faster convergence to Euler’s constant. Math. Gaz. 83, 487–489 (1999)

    Article  Google Scholar 

  10. Tenenbaum, G.: Introduction to Analytic and Probabilistic Number Theory. Cambridge Univ. Press, Cambridge (1995)

    MATH  Google Scholar 

Download references

Acknowledgements

Not applicable.

Availability of data and materials

Not applicable.

Funding

No funding received.

Author information

Authors and Affiliations

Authors

Contributions

The author read and approved the final manuscript.

Corresponding author

Correspondence to G. J. O. Jameson.

Ethics declarations

Competing interests

The author declares that he has no competing interests.

Additional information

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Jameson, G.J.O. A logarithmic estimate for harmonic sums and the digamma function, with an application to the Dirichlet divisor problem. J Inequal Appl 2019, 151 (2019). https://doi.org/10.1186/s13660-019-2104-9

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/s13660-019-2104-9

MSC

  • 26D15
  • 33B15
  • 11N37

Keywords

  • Harmonic sum
  • Euler’s constant
  • Digamma function
  • Divisor problem