On the harmonic number expansion by Ramanujan

Abstract

Let $\gamma =0.577215664\dots$ denote the Euler-Mascheroni constant, and let the sequences

The main aim of this paper is to find the values r, s, t, a, b, c and d which provide the fastest sequences ${(u_n)}_{n\ge 1}$ and ${(v_n)}_{n\ge 1}$ approximating the Euler-Mascheroni constant. Also, we give the upper and lower bounds for ${\sum }_{k=1}^n\frac{1}{k}-\frac{1}{2}ln(n^2+n+\frac{1}{3})-\gamma$ in terms of $n^2+n+\frac{1}{3}$.

MSC: 11Y60, 40A05, 33B15.

1 Introduction

The Euler-Mascheroni constant $\gamma =0.577215664\dots$ is defined as the limit of the sequence

$$D_n=H_n-lnn,$$

(1.1)

where $H_n$ denotes the n th harmonic number defined for $n\in \mathbb{N}:=\{1,2,3,\dots \}$ by

$$H_n=\sum _{k=1}^n\frac{1}{k}.$$

Several bounds for $D_n-\gamma$ have been given in the literature [1–7]. For example, the following bounds for $D_n-\gamma$ were established in [3, 7]:

$$\frac{1}{2(n+1)}<D_n-\gamma <\frac{1}{2n}(n\in \mathbb{N}).$$

The convergence of the sequence $D_n$ to γ is very slow. Some quicker approximations to the Euler-Mascheroni constant were established in [8–21]. For example, Cesàro [8] proved that for every positive integer $n\ge 1$, there exists a number $c_n\in (0,1)$ such that the following approximation is valid:

$$\sum _{k=1}^n\frac{1}{k}-\frac{1}{2}ln(n^2+n)-\gamma =\frac{c_n}{6n(n+1)}.$$

Entry 9 of Chapter 38 of Berndt’s edition of Ramanujan’s Notebooks [[22], p.521] reads,

‘Let $m:=\frac{n(n+1)}{2}$, where n is a positive integer. Then, as n approaches infinity,

$$\begin{array}{rcl}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}& \sim & \frac{1}{2}ln(2m)+\gamma +\frac{1}{12m}-\frac{1}{120m^2}+\frac{1}{630m^3}-\frac{1}{1,680m^4}+\frac{1}{2,310m^5}\\ -\frac{191}{360,360m^6}+\frac{1}{30,030m^7}-\frac{2,833}{1,166,880m^8}+\frac{140,051}{17,459,442m^9}-[\cdots ].\text{’}\end{array}$$

For the history and the development of Ramanujan’s formula, see [20].

Recently, by changing the logarithmic term in (1.1), DeTemple [15], Negoi [18] and Chen et al. [14] have presented, respectively, faster and faster asymptotic formulas as follows:

(1.2)

(1.3)

(1.4)

Chen and Mortici [13] provided a faster asymptotic formula than those in (1.2) to (1.4),

$$\sum _{k=1}^n\frac{1}{k}-ln(n+\frac{1}{2}+\frac{1}{24n}-\frac{1}{48n^2}+\frac{23}{5,760n^3})=\gamma +O\left(n^{-5}\right)(n\to \mathrm{\infty }),$$

(1.5)

and posed the following natural question.

Open problem For a given positive integer p, find the constants $a_j$ ($j=0,1,2,\dots ,p$) such that

$$\sum _{k=1}^n\frac{1}{k}-ln(n+\sum _{j=0}^p\frac{a_j}{n^j})$$

(1.6)

is the sequence which would converge to γ in the fastest way.

Very recently, Yang [21] published the solution of the open problem (1.6) by using logarithmic-type Bell polynomials.

For all $n\in \mathbb{N}$, let

$$P_n=\sum _{k=1}^n\frac{1}{k}-\frac{1}{2}ln(n^2+n+\frac{1}{3})$$

(1.7)

and

$$Q_n=\sum _{k=1}^n\frac{1}{k}-\frac{1}{4}ln[{(n^2+n+\frac{1}{3})}^2-\frac{1}{45}].$$

Chen and Li [12] proved that for all integers $n\ge 1$,

$$\frac{1}{180{(n+1)}^4}<\gamma -P_n<\frac{1}{180n^4}$$

(1.8)

and

$$\frac{8}{2,835{(n+1)}^6}<Q_n-\gamma <\frac{8}{2,835n^6}.$$

Now we define the sequences

$$u_n=\sum _{k=1}^n\frac{1}{k}-\frac{1}{2}ln(n^2+n+\frac{1}{3})-\frac{1}{r(n^2+n+\frac{1}{3})+s{(n^2+n+\frac{1}{3})}^2+t}$$

(1.9)

and

$$\begin{array}{rcl}{v}_n& =& \sum _{k=1}^n\frac{1}{k}-\frac{1}{2}ln(n^2+n+\frac{1}{3})\\ -(\frac{a}{{(n^2+n+\frac{1}{3})}^2}+\frac{b}{{(n^2+n+\frac{1}{3})}^3}+\frac{c}{{(n^2+n+\frac{1}{3})}^4}+\frac{d}{{(n^2+n+\frac{1}{3})}^5}),\end{array}$$

(1.10)

respectively. Our Theorems 1 and 2 are to find the values r, s, t, a, b, c and d which provide the fastest sequences ${(u_n)}_{n\ge 1}$ and ${(v_n)}_{n\ge 1}$ approximating the Euler-Mascheroni constant.

Theorem 1 Let ${(u_n)}_{n\ge 1}$ be defined by (1.9). For

$$r=-\frac{640}{7},s=-180,t=\frac{26,770}{441},$$

we have

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{11}(u_n-u_{n+1})=\frac{457,528}{123,773,265}$$

(1.11)

and

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{10}(u_n-\gamma )=\frac{457,528}{123,773,265}.$$

(1.12)

The speed of convergence of the sequence ${(u_n)}_{n\ge 1}$ is $n^{-10}$.

Theorem 2 Let ${(v_n)}_{n\ge 1}$ be defined by (1.10). For

$$a=-\frac{1}{180},b=\frac{8}{2,835},c=-\frac{5}{1,512},d=\frac{592}{93,555},$$

we have

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{13}(v_n-v_{n+1})=-\frac{796,801}{3,648,645}\mathit{\text{and}}\underset{n\to \mathrm{\infty }}{lim}{n}^{12}(v_n-\gamma )=-\frac{796,801}{43,783,740}.$$

The speed of convergence of the sequence ${(v_n)}_{n\ge 1}$ is $n^{-12}$.

Our Theorems 3 and 4 establish the bounds for $\gamma -P_n$ in terms of $n^2+n+\frac{1}{3}$.

Theorem 3 Let $P_n$ be defined by (1.7). Then

(1.13)

Theorem 4 Let $P_n$ be defined by (1.7). Then

(1.14)

Remark 1 The inequality (1.14) is sharper than (1.8), while the inequality (1.13) is sharper than (1.14).

2 Lemmas

Before we prove the main theorems, let us give some preliminary results.

The constant γ is deeply related to the gamma function $\mathrm{\Gamma }(z)$ thanks to the Weierstrass formula:

$$\mathrm{\Gamma }(z)=\frac{e^{-\gamma z}}{z}\prod _{k=1}^{\mathrm{\infty }}\left\{{(1+\frac{z}{k})}^{-1}{e}^{z/k}\right\}(z\in \mathbb{C}\setminus Z_0^{-};Z_0^{-}:=\{-1,-2,-3,\dots \}).$$

The logarithmic derivative of the gamma function

$$\psi (z)=\frac{{\mathrm{\Gamma }}^{\mathrm{\prime }}(z)}{\mathrm{\Gamma }(z)}\text{or}ln\mathrm{\Gamma }(z)={\int }_1^z\psi (t)\mathrm{d}t$$

is known as the psi (or digamma) function. The successive derivatives of the psi function $\psi (z)$

$${\psi }^{(n)}(z):=\frac{{\mathrm{d}}^n}{\mathrm{d}{z}^n}\{\psi (z)\}(n\in \mathbb{N})$$

are called the polygamma functions.

The following recurrence and asymptotic formulas are well known for the psi function:

$$\psi (z+1)=\psi (z)+\frac{1}{z}$$

(2.1)

(see [[23], p.258]), and

$$\psi (z)\sim lnz-\frac{1}{2z}-\frac{1}{12z^2}+\frac{1}{120z^4}-\frac{1}{252z^6}+\cdots (z\to \mathrm{\infty }\text{ in }|argz|<\pi )$$

(2.2)

(see [[23], p.259]). From (2.1) and (2.2), we get

$$\psi (n+1)\sim lnn+\frac{1}{2n}-\frac{1}{12n^2}+\frac{1}{120n^4}-\frac{1}{252n^6}+\cdots (n\to \mathrm{\infty }).$$

(2.3)

It is also known [[23], p.258] that

$$\psi (n+1)=-\gamma +\sum _{k=1}^n\frac{1}{k}.$$

Lemma 1 [24, 25]

If ${({\lambda }_n)}_{n\ge 1}$ is convergent to zero and there exists the limit

$$\underset{n\to \mathrm{\infty }}{lim}{n}^k({\lambda }_n-{\lambda }_{n+1})=l\in \mathbb{R},$$

with $k>1$, then there exists the limit

$$\underset{n\to \mathrm{\infty }}{lim}{n}^{k-1}{\lambda }_n=\frac{l}{k-1}.$$

Lemma 1 gives a method for measuring the speed of convergence.

Lemma 2 [[26], Theorem 9]

Let $k\ge 1$ and $n\ge 0$ be integers. Then, for all real numbers $x>0$,

$$S_k(2n;x)<{(-1)}^{k+1}{\psi }^{(k)}(x)<S_k(2n+1;x),$$

(2.4)

where

$$S_k(p;x)=\frac{(k-1)!}{x^k}+\frac{k!}{2x^{k+1}}+\sum _{i=1}^p[B_{2i}\prod _{j=1}^{k-1}(2i+j)]\frac{1}{x^{2i+k}},$$

and $B_i$ ($i=0,1,2,\dots$) are Bernoulli numbers defined by

$$\frac{t}{e^t-1}=\sum _{i=0}^{\mathrm{\infty }}{B}_i\frac{t^i}{i!}.$$

It follows from (2.4) that for $x>0$,

$$\begin{array}{r}\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}-\frac{1}{30x^5}+\frac{1}{42x^7}-\frac{1}{30x^9}+\frac{5}{66x^{11}}-\frac{691}{2,730x^{13}}\\ <{\psi }^{\mathrm{\prime }}(x)<\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{6x^3}-\frac{1}{30x^5}+\frac{1}{42x^7}-\frac{1}{30x^9}+\frac{5}{66x^{11}}-\frac{691}{2,730x^{13}}+\frac{7}{6x^{15}},\end{array}$$

from which we imply that for $x>0$,

(2.5)

3 Proofs of Theorems 1-4

Proof of Theorem 1 By using the Maple software, we write the difference $u_n-u_{n+1}$ as a power series in $n^{-1}$:

$$\begin{array}{rcl}{u}_n-u_{n+1}& =& (-\frac{s+180}{45s})\frac{1}{n^5}+\left(\frac{s+180}{9s}\right)\frac{1}{n^6}\\ +\left(\frac{2(-6,048s+567r-32s^2)}{189s^2}\right)\frac{1}{n^7}\\ +\left(\frac{2(-567r+2,268s+11s^2)}{27s^2}\right)\frac{1}{n^8}\\ +\left(\frac{2(-23s^3+2,430sr-5,310s^2+108st-108r^2)}{27s^3}\right)\frac{1}{n^9}\\ +\left(\frac{2(-13,770sr+19,170s^2-1,620st+1,620r^2+73s^3)}{45s^3}\right)\frac{1}{n^{10}}\\ +\frac{1}{2,673s^4}(-15,443s^4+4,834,566s^{2}r-4,650,624s^3+1,033,560s^{2}t\\ -1,033,560s{r}^2-53,460srt+26,730r^3)\frac{1}{n^{11}}\\ +O\left(\frac{1}{n^{12}}\right).\end{array}$$

(3.1)

According to Lemma 1, we have three parameters r, s and t which produce the fastest convergence of the sequence from (3.1)

$$\{\begin{array}{c}s+180=0,\hfill \\ -6,048s+567r-32s^2=0,\hfill \\ -23s^3+2,430sr-5,310s^2+108st-108r^2=0,\hfill \end{array}$$

namely if

$$r=-\frac{640}{7},s=-180,t=\frac{26,770}{441}.$$

Thus, we have

$$u_n-u_{n+1}=\frac{457,528}{123,773,265n^{11}}+O\left(\frac{1}{n^{12}}\right).$$

By using Lemma 1, we obtain the assertion of Theorem 1. □

Proof of Theorem 2 By using the Maple software, we write the difference $v_n-v_{n+1}$ as a power series in $n^{-1}$:

$$\begin{array}{rcl}{v}_n-v_{n+1}& =& (-\frac{1}{45}-4a)\frac{1}{n^5}+(\frac{1}{9}+20a)\frac{1}{n^6}+(-64a-6b-\frac{64}{189})\frac{1}{n^7}\\ +(\frac{22}{27}+168a+42b)\frac{1}{n^8}+(-\frac{1,180}{3}a-8c-\frac{46}{27}-180b)\frac{1}{n^9}\\ +(72c+\frac{146}{45}+852a+612b)\frac{1}{n^{10}}\\ +(-\frac{1,160}{3}c-\frac{15,443}{2,673}-\frac{5,426}{3}b-10d-\frac{46,976}{27}a)\frac{1}{n^{11}}\\ +(\frac{2,375}{243}+\frac{14,542}{3}b+\frac{4,840}{3}c+\frac{91,432}{27}a+110d)\frac{1}{n^{12}}+O\left(\frac{1}{n^{13}}\right).\end{array}$$

(3.2)

According to Lemma 1, we have four parameters a, b, c and d which produce the fastest convergence of the sequence from (3.2)

$$\{\begin{array}{c}-\frac{1}{45}-4a=0,\hfill \\ -64a-6b-\frac{64}{189}=0,\hfill \\ -\frac{1,180}{3}a-8c-\frac{46}{27}-180b=0,\hfill \\ -\frac{1,160}{3}c-\frac{15,443}{2,673}-\frac{5,426}{3}b-10d-\frac{46,976}{27}a=0,\hfill \end{array}$$

namely if

$$a=-\frac{1}{180},b=\frac{8}{2,835},c=-\frac{5}{1,512},d=\frac{592}{93,555}.$$

Thus, we have

$$v_n-v_{n+1}=-\frac{796,801}{3,648,645n^{13}}+O\left(\frac{1}{n^{14}}\right).$$

By using Lemma 1, we obtain the assertion of Theorem 2. □

Proof of Theorem 3 Here we only prove the second inequality in (1.13). The proof of the first inequality in (1.13) is similar. The upper bound of (1.13) is obtained by considering the function F for $x\ge 1$ defined by

$$F(x)=\frac{1}{2}ln(x^2+x+\frac{1}{3})-\psi (x+1)-\frac{1}{\frac{640}{7}(n^2+n+\frac{1}{3})+180{(n^2+n+\frac{1}{3})}^2-\frac{26,770}{441}}.$$

Differentiation and applying the right-hand inequality of (2.5) yield

$$\begin{array}{rcl}{F}^{\mathrm{\prime }}(x)& =& -{\psi }^{\mathrm{\prime }}(x+1)+\frac{2x+1}{2(x^2+x+\frac{1}{3})}\\ +\frac{55,566(126x^3+189x^2+137x+37)}{5{(7,938x^4+15,876x^3+17,262x^2+9,324x-451)}^2}\\ >& -(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{6x^3}-\frac{1}{30x^5}+\frac{1}{42x^7}-\frac{1}{30x^9}+\frac{5}{66x^{11}}-\frac{691}{2,730x^{13}}+\frac{7}{6x^{15}})\\ +\frac{2x+1}{2(x^2+x+\frac{1}{3})}+\frac{55,566(126x^3+189x^2+137x+37)}{5{(7,938x^4+15,876x^3+17,262x^2+9,324x-451)}^2}\\ =& \frac{P(x)}{30,030x^{13}{(3x^2+3x+1)}^6},\end{array}$$

where

$$\begin{array}{rcl}P(x)& =& 35,471,898,974,548,627,145+138,773,138,144,376,345,519(x-4)\\ +241,909,257,272,859,643,240{(x-4)}^2\\ +253,899,751,881,744,791,655{(x-4)}^3\\ +181,059,030,163,487,870,836{(x-4)}^4\\ +93,303,260,620,236,720,571{(x-4)}^5\\ +35,932,291,146,874,735,228{(x-4)}^6+10,519,794,292,714,982,599{(x-4)}^7\\ +2,353,926,972,956,528,576{(x-4)}^8+400,626,844,002,342,775{(x-4)}^9\\ +51,041,813,866,867,916{(x-4)}^{10}+4,719,218,347,433,667{(x-4)}^{11}\\ +299,247,577,164,158{(x-4)}^{12}+11,646,155,626,560{(x-4)}^{13}\\ +209,840,641,920{(x-4)}^{14}>0\text{for }x\ge 4.\end{array}$$

Therefore, $F^{\mathrm{\prime }}(x)>0$ for $x\ge 4$.

For $x=1,2,3,4$, we compute directly:

Hence, the sequence ${(F(n))}_{n\ge 1}$ is strictly increasing. This leads to

$$F(n)<\underset{n\to \mathrm{\infty }}{lim}F(n)=0$$

by using the asymptotic formula (2.3). This completes the proof of the second inequality in (1.13). □

Proof of Theorem 4 Here we only prove the first inequality in (1.14). The proof of the second inequality in (1.14) is similar. The lower bound of (1.14) is obtained by considering the function G for $x\ge 1$ defined by

$$\begin{array}{rcl}G(x)& =& \psi (x+1)-\frac{1}{2}ln(x^2+x+\frac{1}{3})\\ +(\frac{\frac{1}{180}}{{(x^2+x+\frac{1}{3})}^2}+\frac{-\frac{8}{2,835}}{{(x^2+x+\frac{1}{3})}^3}+\frac{\frac{5}{1,512}}{{(x^2+x+\frac{1}{3})}^4}+\frac{-\frac{592}{93,555}}{{(x^2+x+\frac{1}{3})}^5}).\end{array}$$

Differentiation and applying the left-hand inequality of (2.5) yield

$$\begin{array}{rcl}{G}^{\mathrm{\prime }}(x)& =& {\psi }^{\mathrm{\prime }}(x+1)-\frac{2x+1}{2(x^2+x+\frac{1}{3})}\\ -\frac{3(4,158x^7+14,553x^6+19,701x^5+12,870x^4+8,283x^3+6,831x^2-8,276x-5,194)}{770{(3x^2+3x+1)}^6}\\ >& (\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{6x^3}-\frac{1}{30x^5}+\frac{1}{42x^7}-\frac{1}{30x^9}+\frac{5}{66x^{11}}-\frac{691}{2,730x^{13}})-\frac{2x+1}{2(x^2+x+\frac{1}{3})}\\ -\frac{3(41,58x^7+14,553x^6+19,701x^5+12,870x^4+8,283x^3+6,831x^2-8,276x-5,194)}{770{(3x^2+3x+1)}^6}\\ =& \frac{Q(x)}{30,030x^{13}{(3x^2+3x+1)}^6},\end{array}$$

where

$$\begin{array}{rcl}Q(x)& =& 274,317,996,839,484+1,074,684,262,984,527(x-5)\\ +1,571,352,927,565,772{(x-5)}^2+1,266,557,271,610,345{(x-5)}^3\\ +652,427,951,634,329{(x-5)}^4+230,639,944,842,034{(x-5)}^5\\ +57,987,546,990,473{(x-5)}^6+10,515,845,175,406{(x-5)}^7\\ +1,371,027,303,124{(x-5)}^8+125,702,024,549{(x-5)}^9\\ +7,709,579,845{(x-5)}^{10}+284,457,957{(x-5)}^{11}\\ +4,780,806{(x-5)}^{12}>0\text{for }x\ge 5.\end{array}$$

Therefore, $G^{\mathrm{\prime }}(x)>0$ for $x\ge 5$.

For $x=1,2,3,4,5$, we compute directly:

Hence, the sequence ${(G(n))}_{n\ge 1}$ is strictly increasing. This leads to

$$G(n)<\underset{n\to \mathrm{\infty }}{lim}G(n)=0$$

by using the asymptotic formula (2.3). This completes the proof of the first inequality in (1.14). □

Remark 2 Some calculations in this work were performed by using the Maple software for symbolic calculations.

Remark 3 The work of the first author was supported by a grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, project number PN-II-ID-PCE-2011-3-0087.