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.

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.

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.

2 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 $\log f(x)$.

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

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

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

 □