Median Bernoulli Numbers and Ramanujan’s Harmonic Number Expansion

Abstract

Ramanujan-type harmonic number expansion was given by many authors. Some of the most well-known are: $H_n\sim \gamma +logn-{\sum }_{k=1}^{\infty }\frac{B_k}{k·n^k}$, where $B_k$ is the Bernoulli numbers. In this paper, we rewrite Ramanujan’s harmonic number expansion into a similar form of Euler’s asymptotic expansion as n approaches infinity: $H_n\sim \gamma +c_0\left(h\right)log(q+h)-{\sum }_{k=1}^{\infty }\frac{c_k\left(h\right)}{k·{(q+h)}^k}$, where $q=n(n+1)$ is the nth pronic number, twice the nth triangular number, $\gamma$ is the Euler–Mascheroni constant, and $c_k\left(x\right)={\sum }_{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)c_j{x}^{k-j}$, with $c_k$ is the negative of the median Bernoulli numbers. Then, $2c_n={\sum }_{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n+k}$, where $B_n$ is the Bernoulli number. By using the result obtained, we present two general Ramanujan’s asymptotic expansions for the nth harmonic number. For example, $H_n\sim \gamma +\frac{1}{2}log(q+\frac{1}{3})-\frac{1}{180{(q+\frac{1}{3})}^2}{\left({\sum }_{j=0}^{\infty }\frac{b_j\left(r\right)}{{(q+\frac{1}{3})}^j}\right)}^{1/r}$ as n approaches infinity, where $b_j\left(r\right)$ can be determined.

1. Introduction

Leonhard Euler in 1755 applied the Euler–Maclaurin sum formula to find the famous standard Euler asymptotic expansion for $H_n$ as $n\to \infty$:

$$H_n\sim \gamma +logn-\sum _{k=1}^{\infty }\frac{B_k}{k·n^k},$$

(1)

where $B_k$ is the Bernoulli number defined by $\frac{t}{e^t-1}={\sum }_{k=0}^{\infty }\frac{B_k}{k!}{t}^k$, and $\gamma =0.57721\cdots$ is the Euler–Mascheroni constant.

Ramanujan [1] proposed the following asymptotic expansion for $H_n$:

$$\begin{array}{cc}\hfill H_n\sim & \gamma +\frac{1}{2}log\left(2m\right)+\frac{1}{12m}-\frac{1}{120m^2}+\frac{1}{630m^3}-\frac{1}{1680m^4}+\frac{1}{2310m^5}\hfill \\ \hfill & -\frac{191}{360360m^6}+\frac{29}{30030m^7}-\frac{2833}{1166880m^8}+\frac{140051}{17459442m^9}-\cdots ,\hfill \end{array}$$

(2)

where $m=n(n+1)/2$ is the n-th triangular number. However, Ramanujan did not give any formulas for the general terms and also without any proof. Rewrite the above formula as the following notation:

$$H_n\sim \gamma +\frac{1}{2}log\left(2m\right)+\sum _{k=1}^{\infty }\frac{R_k}{m^k}.$$

(3)

In 2008, Villarino [2] established an explicit expression for the coefficient sequence $\left(R_k\right)$:

$$R_k=\frac{{(-1)}^{k-1}}{2k·8^k}\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right){(-4)}^j{B}_{2j}(1/2),$$

(4)

where $B_k\left(x\right)$ are the Bernoulli polynomials defined by $\frac{t{e}^{xt}}{e^t-1}={\sum }_{n=0}^{\infty }\frac{B_n\left(x\right)t^n}{n!}$. In 2015, Chen and Cheng [3] reconsidered Ramanujan’s formula and gave the following recurrence relation for $\left(R_k\right)$:

$$R_1=\frac{1}{12},R_k=\frac{1}{2^k}\left\{\frac{1}{4k}-\frac{B_{2k}}{2k}-\sum _{j=1}^{k-1}{2}^j{R}_j\left(\genfrac{}{}{0pt}{}{2k-j-1}{j-1}\right)\right\},k\ge 2.$$

(5)

In 2019, Chen [4] improved the recurrence relation as

$$R_k=\frac{1}{2^{k+1}k}\left\{\frac{1}{2k+1}-\sum _{j=1}^{k-1}{2}^{j+1}{R}_j\left(\genfrac{}{}{0pt}{}{2k-j}{2k-2j+1}\right)\right\},\mathrm{for} k\ge 2.$$

(6)

Another Ramanujan-type harmonic number expansion was given by Wang [5] in 2018,

$$H_n\sim \gamma +\frac{1}{2}log(2m+h)+\sum _{k=1}^{\infty }\frac{{\alpha }_k\left(h\right)}{{(2m+h)}^k},$$

(7)

where h is a parameter and $\left({\alpha }_k\left(h\right)\right)$ is a coefficient sequence

$${\alpha }_k\left(h\right)=-\frac{h^k}{2k}+\sum _{j=1}^k\left(\genfrac{}{}{0pt}{}{k-1}{j-1}\right)R_j{2}^j{h}^{k-j}.$$

(8)

In this paper, we rewrite Ramanujan’s harmonic number expansion into a similar form of Euler’s asymptotic expansion:

$$H_n\sim \gamma +c_{0}logq-\sum _{k=1}^{\infty }\frac{c_k}{k·q^k},$$

(9)

where $q=n(n+1)=2m$ is the nth pronic number, twice the nth triangular number. In fact, we prove that the number $c_k$ is the negative of the median Bernoulli number. The median Bernoulli number is studied by the author [6] in 2005. Then, we have for $n\ge 0$,

$$c_n=\frac{1}{2}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n+k}.$$

(10)

Moreover, let

$$c_k\left(x\right)=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)c_j{x}^{k-j}.$$

(11)

Then, we could rewrite Wang’s expansion Equation (7) as follows:

$$H_n\sim \gamma +c_0\left(h\right)log(q+h)-\sum _{k=1}^{\infty }\frac{c_k\left(h\right)}{k·{(q+h)}^k}.$$

(12)

We give simpler asymptotic expansion representations for $H_n$ using Equations (10) and (11), which effectively integrate the results of Villarino, Chen and Cheng, Chen, and Wang (see Equations (4)–(7)) and make their representations more meaningful. We discuss some properties of the numbers $c_n$ and the polynomials $c_n\left(x\right)$ in Section 3 and Section 4, respectively. For example, the Hankel determinant of $c_n\left(x\right)$ for any x can be evaluated as

$$2^{n+1}\underset{0\le i,j\le n}{det}\left(c_{i+j}\left(x\right)\right)=\underset{0\le i,j\le n}{det}\left(B_{2i+2j}(1/2)\right).$$

(13)

Furthermore, Chen [7] gave a new asymptotic expansion. For any nonzero real number r, the n-th harmonic number $H_n$ may have an asymptotic expansion as n approaches infinity:

$$\frac{1}{2}log\left(2m\right)+\gamma +\frac{1}{12m}{\left(\sum _{j=0}^{\infty }\frac{a_j\left(r\right)}{m^j}\right)}^{1/r},$$

(14)

where the parameters $a_j\left(r\right)$ satisfy the following recurrence relation

$$a_0\left(r\right)=1,a_j\left(r\right)=\frac{1}{j}\sum _{k=1}^j\left[k(1+r)-j\right]\left(12R_{k+1}\right)a_{j-k}\left(r\right),j\in \mathbb{N}.$$

Inspired by this, we give a more general asymptotic expansion in Section 5 using Equation (12). Given $r,h$ real numbers with $r\ne 0$, $h\ne 1/3$, we get

$$H_n\sim \gamma +c_0\left(h\right)log(q+h)-\frac{3h-1}{6(q+h)}{\left(1+\sum _{j=1}^{\infty }\frac{a_j(r,h)}{{(q+h)}^j}\right)}^{1/r},n\to \infty ,$$

(15)

We know that the formula with $h=0$ is Equation (14) (see ([7], Theorem 2.3)). Since $c_1\left(h\right)=\frac{3h-1}{6}$, $h=1/3$ will remove the ${(q+h)}^{-1}$ term. This will improve the approximation. Thus, it can be seen that there are a lot of investigations for the $h=1/3$ case, see [4,8,9,10].

If $h=1/3$, then the asymptotic expansion will become

$$H_n\sim \gamma +\frac{1}{2}log\left(q+\frac{1}{3}\right)-\frac{1}{180{\left(q+\frac{1}{3}\right)}^2}{\left(1+\sum _{j=1}^{\infty }\frac{b_j\left(r\right)}{{\left(q+\frac{1}{3}\right)}^j}\right)}^{1/r}.$$

(16)

The parameters $a_j(r,h)$ and $b_j\left(r\right)$ in Equations (15) and (16) are determined by some recurrence relations, which will be illustrated in Theorems 2 and 3, respectively. At the end of this paper, we will compare how close these asymptotic formulas are to $H_n$.

2. Median Bernoulli Numbers and $R_k$

Set $a_{0,n}=B_n$, for $n\ge 0$. And for $n\ge 1,k\ge 0$,

$$a_{n,k}=a_{n-1,k}+a_{n-1,k+1},$$

or equivalently,

$$a_{n,k}=\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right)a_{0,k+j}.$$

The corresponding matrix is represented as follows.

$$\begin{array}{cccccccc}\hfill 1& \hfill -1/2& \hfill 1/6& \hfill 0& \hfill -1/30& \hfill 0& \hfill 1/42& \hfill 0\\ \hfill 1/2& \hfill -1/3& \hfill 1/6& \hfill -1/30& \hfill -1/30& \hfill 1/42& \hfill 1/42& \hfill \cdots \\ \hfill 1/6& \hfill -1/6& \hfill 2/15& \hfill -1/15& \hfill -1/105& \hfill 1/21& \hfill \cdots \\ \hfill 0& \hfill -1/30& \hfill 1/15& \hfill -8/105& \hfill 4/105& \hfill \cdots \\ \hfill -1/30& \hfill 1/30& \hfill -1/105& \hfill -4/105& \hfill \cdots \\ \hfill 0& \hfill 1/42& \hfill -1/21& \hfill \cdots \\ \hfill 1/42& \hfill -1/42& \hfill \cdots \\ \hfill 0& \hfill \cdots \\ \hfill \cdots \end{array}$$

This matrix is called the “$BS$-matrix” in [6], which is a special Euler–Seidel matrix. Let

$$c_n=a_{n+1,n}$$

be the lower diagonal sequence of the $BS$-matrix. The number $c_n$ is the negative of the median Bernoulli number $K_n$, which is the upper diagonal sequence of the $BS$-matrix, i.e., $K_n=a_{n,n+1}=-c_n$ (ref. [6]). Therefore, by [6], and Equations (8), (15) and (16), we have

$$c_n=\frac{1}{2}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n+k}=\sum _{k=0}^{n+1}\left(\genfrac{}{}{0pt}{}{n+1}{k}\right)B_{n+k}=-\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n+1+k}.$$

(17)

Let the ordinary generating function of $c_n$ as follows.

$$m\left(x\right)=\sum _{n=0}^{\infty }{c}_n{x}^{n+1}.$$

Let $\psi \left(x\right)$ be the formal Laplace transform of $t/sinht$. Then, the following relation was obtained ([6], Theorem 4.2, Equation (29))

$$2x·\psi \left(x\right)=m\left(\frac{4x^2}{1-x^2}\right).$$

(18)

Since $t/sinht={\sum }_{n=0}^{\infty }{4}^n{B}_{2n}(1/2)t^{2n}/\left(2n\right)!$, we have that for $n\ge 0$ ([6], Equation (32)),

$$2^{2n+1}{(-1)}^n{c}_n=\sum _{j=0}^n\left(\genfrac{}{}{0pt}{}{n}{j}\right){(-1)}^j{2}^{2j}{B}_{2j}(1/2).$$

(19)

Using Villarino’s explicit formula for $R_k$, Equation (4), we have for $k\ge 1$,

$$-2^k·k·R_k=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right){(-1)}^{k+j}{2}^{2j-2k-1}{B}_{2j}(1/2)=c_k.$$

(20)

This implies that

$$H_n\sim \gamma +c_{0}logq-\sum _{k=1}^{\infty }\frac{c_k}{k{q}^k},n\to \infty .$$

(21)

On the other hand, we substitute $R_k$ as Equation (20) in Wang’s formula for ${\alpha }_k\left(h\right)$, (see Equation (8)), we have for $k\ge 1$,

$$\begin{array}{cc}\hfill -k·{\alpha }_k\left(h\right)& =\frac{h^k}{2}-\sum _{j=1}^k\left(\genfrac{}{}{0pt}{}{k-1}{j-1}\right)R_j{2}^k{h}^{k-j}k\hfill \\ \hfill & =\frac{h^k}{2}+\sum _{j=1}^k\left(\genfrac{}{}{0pt}{}{k-1}{j-1}\right)h^{k-j}\frac{c_j}{j}·k=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)h^{k-j}{c}_j=c_k\left(h\right).\hfill \end{array}$$

(22)

Therefore, we conclude our result in the following.

Theorem 1.

For $n\to \infty$, we have

$$H_n\sim \gamma +c_0\left(h\right)log(q+h)-\sum _{k=1}^{\infty }\frac{c_k\left(h\right)}{k·{(q+h)}^k},$$

(23)

where $q=n(n+1)$ is the nth pronic number,

$$c_k\left(x\right)=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)c_j{x}^{k-j},and{c}_n=\frac{1}{2}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)B_{n+k}.$$

(24)

3. Some Properties of $c_n$

Let the ordinary generating function of $B_n$ as follows.

$$b\left(x\right)=\sum _{n=0}^{\infty }{B}_n{x}^{n+1}.$$

Using the relation between the ordinary generating functions of $a_{0,n}$, $a_{n,n}$, and $a_{n,n+1}$ of the $BS$-matrix, we have the following relation ([6], Theorem 4.2, Equation (29))

$$b\left(x\right)=\left(1+\frac{2}{x}\right)m\left(\frac{x^2}{1+x}\right).$$

(25)

Then, the following identity is obtained ([6], Equation (27)).

$$\sum _{j=0}^{\lfloor n/2\rfloor }\left(\genfrac{}{}{0pt}{}{n-j}{j}\right)\frac{n}{n-j}{c}_j=\left\{\begin{array}{cc}\hfill -B_1,& \hfill \mathrm{if}n=1,\\ \hfill B_n,& \hfill \mathrm{if}n\ge 2.\end{array}\right.$$

(26)

In the above formula, the formula obtained by substituting $n=2k$ appears in the recurrence relation of $R_k$ given by Chen and Cheng [3] in 2015 (see Equation (5)). Furthermore, if we substitute $n=2k+1$ into the above identity, we obtain Equation (6) given by Chen [4] in 2019.

There are a lot of properties of $c_n$ obtained from [6]. For example, let the denominators and the numerators of the rational number $c_n$ be $D_n,N_n$, respectively. We have the following properties ([6], Theorem 1.1):

• The denominator $D_n$ is a square-free integer.
• The set of the all odd prime divisors of $D_n$ is
  $\left\{p:\mathrm{odd}\mathrm{prime}\mid \frac{n}{m}\le p-1\le \frac{2n}{2m-1},m\in \mathbb{N}\right\}$.
• The denominator $D_n$ is an odd integer, for $n\ge 2$.
• The largest power of 2 that divides the numerator $N_n$ is $2\lfloor \frac{n-1}{2}\rfloor$.

The ordinary generating function $m\left(x\right)$ has the following continued fraction representation ([6], Theorem 5.5)

$$m\left(x\right)=\frac{c_{0}x}{1+a_{0}x}-\frac{b_1{x}^2}{1+a_{1}x}-\frac{b_2{x}^2}{1+a_{2}x}-\phantom{+}\cdots ,$$

(27)

where for $n\ge 0$,

$$\begin{array}{cc}\hfill a_n& =\frac{8n^4+8n^3+6n^2+2n-1}{(4n+3)(4n-1)},\hfill \\ \hfill b_{n+1}& =\frac{{(2n+1)}^4{(n+1)}^4}{(4n+1){(4n+3)}^2(4n+5)}.\hfill \end{array}$$

Using this representation, we have the Hankel determinant of $c_n$ (ref. [6], Theorem 5.5)

$$\underset{0\le i,j\le n}{det}\left(c_{i+j}\right)={\left(\frac{1}{2}\right)}^{n+1}\prod _{j=1}^n{\left(\frac{{(2j-1)}^4{j}^4}{(4j-3){(4j-1)}^2(4j+1)}\right)}^{n-j+1}.$$

(28)

Since the finite product in Equation (28) is the Hankel determinant of $B_{2n}(1/2)$ (see [6], Equation (41)), we have

$$\underset{0\le i,j\le n}{det}\left(B_{2i+2j}(1/2)\right)=2^{n+1}\underset{0\le i,j\le n}{det}\left(c_{i+j}\right).$$

(29)

By Equation (19) and an integral representation of $B_{2n}(1/2)$ ([11], Equation (28))

$$B_{2n}(1/2)={(-1)}^n\pi {\int }_0^{\infty }{t}^{2n}{sech}^2\left(\pi t\right)dt,$$

we have an integral representation of $c_n$, for $n\ge 0$,

$$c_n=\frac{{(-1)}^n\pi }{2^{2n+1}}{\int }_0^{\infty }{(4t^2+1)}^n{sech}^2\left(\pi t\right)dt.$$

(30)

4. Some Properties of ${\mathit{c}}_{\mathit{n}}\left(\mathit{x}\right)$

We first list $c_n\left(x\right)$ for $n=0,1,2,\dots ,5$.

$$\begin{array}{cc}\hfill c_0\left(x\right)& =\frac{1}{2},\hfill \\ \hfill c_1\left(x\right)& =-\frac{1}{6}+\frac{x}{2},\hfill \\ \hfill c_2\left(x\right)& =\frac{1}{15}-\frac{x}{3}+\frac{x^2}{2},\hfill \\ \hfill c_3\left(x\right)& =-\frac{4}{105}+\frac{x}{5}-\frac{x^2}{2}+\frac{x^3}{2},\hfill \\ \hfill c_4\left(x\right)& =\frac{4}{105}-\frac{16x}{105}+\frac{2x^2}{5}-\frac{2x^3}{3}+\frac{x^4}{2},\hfill \\ \hfill c_5\left(x\right)& =-\frac{16}{231}+\frac{4x}{21}-\frac{8x^2}{21}+\frac{2x^3}{3}-\frac{5x^4}{6}+\frac{x^5}{2}.\hfill \end{array}$$

Differentiating Equation (11) with respect to x we obtain

$$\begin{array}{cc}\hfill \frac{d}{dx}{c}_n\left(x\right)& =\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)c_{n-k}k{x}^{k-1}\hfill \\ \hfill & =n\sum _{k=0}^{n-1}\left(\genfrac{}{}{0pt}{}{n-1}{k}\right)c_{n-1-k}{x}^k=n{c}_{n-1}\left(x\right).\hfill \end{array}$$

(31)

Therefore,

$${\int }_x^y{c}_n\left(t\right)dt=\frac{c_{n+1}\left(y\right)-c_{n+1}\left(x\right)}{n+1}.$$

(32)

On the other hand, we use Equation (30) to get an integral representation of $c_n\left(x\right)$:

$$c_n\left(x\right)=\frac{{(-1)}^n\pi }{2^{2n+1}}{\int }_0^{\infty }{(4t^2-4x+1)}^n{sech}^2\left(\pi t\right)dt.$$

(33)

Let us consider the function $c_n(x+y)$. We express ${(x+y)}^k$ as its binomial expansion.

$$c_n(x+y)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)c_{n-k}{(x+y)}^k=\sum _{k=0}^n{c}_{n-k}\sum _{\ell =0}^k\left(\genfrac{}{}{0pt}{}{k}{\ell }\right)x^{\ell }{y}^{k-\ell }.$$

We interchange the order of summation and the inner sum becomes $c_{n-\ell }\left(y\right)$:

$$c_n(x+y)=\sum _{\ell =0}^n\sum _{k=\ell }^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)\left(\genfrac{}{}{0pt}{}{k}{\ell }\right)c_{n-k}{x}^{\ell }{y}^{k-\ell }=\sum _{\ell =0}^n\left(\genfrac{}{}{0pt}{}{n}{\ell }\right)x^{\ell }{c}_{n-\ell }\left(y\right).$$

Thus, we have

$$c_n(x+y)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)c_k\left(x\right)y^{n-k}.$$

(34)

Using the inversion binomial theorem to Equation (19) we have

$$4^n{B}_{2n}(1/2)=\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^{2k+1}{c}_k=2^{2n+1}{c}_n(1/4).$$

This implies that

$$2c_n(1/4)=B_{2n}(1/2).$$

(35)

Applying the above identity and Equation (29), we know that the Hankel determinant of $c_n(1/4)$ is the same as the Hankel determinant of $c_n$. However, we use ([12], Proposition 1), indeed that for any value of x,

$$\underset{0\le i,j\le n}{det}\left(c_{i+j}\right)=\underset{0\le i,j\le n}{det}\left(c_{i+j}\left(x\right)\right).$$

(36)

5. New Asymptotic Expansions

To derive our new asymptotic expansions are inspired by ([7], Theorem 2.3). We need the following lemma.

Lemma 1

([7], Lemma 1). If ${\sum }_{j=0}^{\infty }{q}_j{x}^{-j}$ is an asymptotic expansion for $g\left(x\right)$ as x approaches infinity. Given any real number r, the parameters $Q_j\left(r\right)$ are defined by $Q_0\left(r\right)=1$ and for $j\in \mathbb{N}$,

$$Q_j\left(r\right)=\frac{1}{j}\sum _{k=1}^j\left(k(1+r)-j\right)q_k{Q}_{j-k}\left(r\right).$$

Then ${\sum }_{j=0}^{\infty }{Q}_j\left(r\right)x^{-j}$ is an asymptotic expansion for $g{\left(x\right)}^r$.

Our new asymptotic expansions are derived from Equation (23). It is note that $c_1\left(h\right)=\frac{3h-1}{6}$. Therefore, we divide into two cases depending on whether h is $1/3$ or not.

Theorem 2.

Let r and h be any given real numbers with $r\ne 0$ and $h\ne 1/3$. The n-th harmonic number $H_n$ has the following asymptotic expansion as n approaches infinity:

$$\gamma +c_0\left(h\right)log(q+h)-\frac{3h-1}{6(q+h)}{\left(1+\sum _{j=1}^{\infty }\frac{a_j(r,h)}{{(q+h)}^j}\right)}^{1/r}$$

(37)

where the parameters $a_j(r,h)$ given by the recurrence relation

$$\begin{array}{cc}\hfill a_0(r,h)& =1,\hfill \\ \hfill a_j(r,h)& =\frac{1}{j}\sum _{k=1}^j\left(k(1+r)-j\right)\frac{6c_{k+1}\left(h\right)}{(3h-1)(k+1)}{a}_{j-k}(r,h),j\ge 1.\hfill \end{array}$$

(38)

Proof. 

Rewrite Equation (37) as the following representation:

$$\frac{6(q+h)}{3h-1}\left(H_n-\gamma -c_0\left(h\right)log(q+h)\right)\sim {\left(1+\sum _{j=1}^{\infty }\frac{a_j(r,h)}{{(q+h)}^j}\right)}^{1/r}.$$

In view of Equation (23), we have

$$\frac{6(q+h)}{3h-1}\left(H_n-\gamma -c_0\left(h\right)log(q+h)\right)\sim \left(1+\frac{6}{3h-1}\sum _{k=1}^{\infty }\frac{c_{k+1}\left(h\right)}{(k+1){(q+h)}^k}\right).$$

Comparing the above two expressions, we know that

$${\left(1+\frac{6}{3h-1}\sum _{k=1}^{\infty }\frac{c_{k+1}\left(h\right)}{(k+1){(q+h)}^k}\right)}^r\sim 1+\sum _{j=1}^{\infty }\frac{a_j(r,h)}{{(q+h)}^j}.$$

We apply Lemma 1 and get the result we want. □

Using a similar approach, we can easily derive the following theorem for the situation $h=1/3$.

Theorem 3.

Given a real number r with $r\ne 0$. The n-th harmonic number $H_n$ has the asymptotic expansion as n approaches infinity:

$$\gamma +\frac{1}{2}log(q+\frac{1}{3})-\frac{1}{180{(q+\frac{1}{3})}^2}{\left(1+\sum _{j=1}^{\infty }\frac{b_j\left(r\right)}{{(q+\frac{1}{3})}^j}\right)}^{1/r}$$

(39)

where the parameters $b_j\left(r\right)$ are defined by the following relation

$$b_0\left(r\right)=1,b_j\left(r\right)=\frac{1}{j}\sum _{k=1}^j\left(k(1+r)-j\right)\frac{180c_{k+2}\left(\frac{1}{3}\right)}{k+2}{b}_{j-k}\left(r\right),j\ge 1.$$

(40)

Chen [7] discussed many properties of the $h=0$ case. Therefore, we mainly deal with the case of $h=1/3$ here.

The first few parameters $b_j\left(r\right)$ are:

$$\begin{array}{cc}\hfill b_0\left(r\right)& =1,\hfill \\ \hfill b_1\left(r\right)& =-\frac{32}{63}r,\hfill \\ \hfill b_2\left(r\right)& =\frac{3701}{7938}r+\frac{512}{3969}{r}^2,\hfill \\ \hfill b_3\left(r\right)& =-\frac{7264240}{8251551}r-\frac{59216}{250047}{r}^2-\frac{16384}{750141}{r}^3,\hfill \\ \hfill b_4\left(r\right)& =\frac{47882328785}{18021387384}r+\frac{2311659673}{4158781704}{r}^2+\frac{947456}{15752961}{r}^3+\frac{131072}{47258883}{r}^4,\hfill \\ \hfill b_5\left(r\right)& =-\frac{8014919889976}{709592128245}r-\frac{749340134980}{425755276947}{r}^2-\frac{789621116}{4678629417}{r}^3\hfill \\ \hfill & -\frac{30318592}{2977309629}{r}^4-\frac{4194304}{14886548145}{r}^5.\hfill \end{array}$$

For $r=1$ in Equation (39), the resulting asymptotic expansion is as follows ([10], Equation (3.24)):

$$\begin{array}{cc}\hfill H_n\sim & \gamma +\frac{1}{2}log(q+\frac{1}{3})-\frac{1}{180{(q+\frac{1}{3})}^2}+\frac{8}{2835{(q+\frac{1}{3})}^3}-\frac{5}{1512{(q+\frac{1}{3})}^4}\hfill \\ \hfill & +\frac{592}{93555{(q+\frac{1}{3})}^5}-\frac{796801}{43783740{(q+\frac{1}{3})}^6}+\frac{268264}{3648645{(q+\frac{1}{3})}^7}-\cdots \hfill \end{array}$$

(41)

as $n\to \infty$.

For $r=-1$ in Equation (39), we obtain a new asymptotic expansion:

$$\begin{array}{cc}\hfill H_n\sim & \gamma +\frac{1}{2}log(q+\frac{1}{3})-\left[180{(q+\frac{1}{3})}^2+\frac{640}{7}(q+\frac{1}{3})-\frac{26770}{441}+\frac{36602240}{305613(q+\frac{1}{3})}\right.\hfill \\ \hfill & {\left.-\frac{97247611025}{250297047{(q+\frac{1}{3})}^2}+\frac{27515011460000}{15768713961{(q+\frac{1}{3})}^3}-\cdots \right]}^{-1}\hfill \end{array}$$

(42)

as $n\to \infty$.

For $r=-23/2$ in Equation (39), we obtain a new asymptotic expansion:

$$\begin{array}{cc}\hfill H_n& \sim \gamma +\frac{1}{2}log(q+\frac{1}{3})-\frac{1}{180{(q+\frac{1}{3})}^2}\left[1+\frac{368}{63(q+\frac{1}{3})}+\frac{185725}{15876{(q+\frac{1}{3})}^2}\right.\hfill \\ \hfill & {\left.+\frac{3674204}{305613{(q+\frac{1}{3})}^3}-\frac{5793677}{728136864{(q+\frac{1}{3})}^4}+\frac{1021070020123}{31537427922{(q+\frac{1}{3})}^5}+\cdots \right]}^{-2/23}\hfill \end{array}$$

(43)

as $n\to \infty$.

From a computational point of view, the formulas Equations (42) and (43) are better than Equation (41).

It follows from Equations (41)–(43) that for $n\to \infty$,

$$\begin{array}{cc}\hfill H_n& \sim \gamma +\frac{1}{2}log(q+\frac{1}{3})\hfill \\ \hfill & -\frac{1}{180{(q+\frac{1}{3})}^2}+\frac{8}{2835{(q+\frac{1}{3})}^3}-\frac{5}{1512{(q+\frac{1}{3})}^4}+\frac{592}{93555{(q+\frac{1}{3})}^5}:=u_n,\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill H_n& \sim \gamma +\frac{1}{2}log(q+\frac{1}{3})\hfill \\ \hfill & -\frac{1}{180{(q+\frac{1}{3})}^2+\frac{640}{7}(q+\frac{1}{3})-\frac{26770}{441}+\frac{36602240}{305613(q+\frac{1}{3})}}:=v_n,\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill H_n& \sim \gamma +\frac{1}{2}log(q+\frac{1}{3})\hfill \\ \hfill & -\frac{1}{180{(q+\frac{1}{3})}^2{\left[1+\frac{368}{63(q+\frac{1}{3})}+\frac{185725}{15876{(q+\frac{1}{3})}^2}+\frac{3674204}{305613{(q+\frac{1}{3})}^3}\right]}^{2/23}}:=w_n.\hfill \end{array}$$

(46)

From Table 1, we observe that, among approximation formulas Equations (44)–(46), for $n\ge 1$, the formula Equation (46) would be the best one. There seems to be an optimal real number r in Equation (39), and when we substitute it in this formula, the resulting approximation should be optimal. We guess that this real number r should be close to $-11.502534\dots$

Table 1. Comparison of approximation Formulas (44)–(46).

n	${\mathit{u}}_{\mathit{n}}-{\mathit{H}}_{\mathit{n}}$	${\mathit{v}}_{\mathit{n}}-{\mathit{H}}_{\mathit{n}}$	${\mathit{w}}_{\mathit{n}}-{\mathit{H}}_{\mathit{n}}$
1	$4.625\times {10}^{-5}$	$1.997\times {10}^{-5}$	$-1.405\times {10}^{-6}$
10	$9.735\times {10}^{-15}$	$6.332\times {10}^{-15}$	$-6.750\times {10}^{-17}$
${10}^2$	$1.713\times {10}^{-26}$	$1.129\times {10}^{-26}$	$2.162\times {10}^{-30}$
${10}^3$	$1.809\times {10}^{-38}$	$1.192\times {10}^{-38}$	$3.805\times {10}^{-42}$
${10}^4$	$1.819\times {10}^{-50}$	$1.198\times {10}^{-50}$	$3.841\times {10}^{-54}$