# 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 ${\left({u}_{n}\right)}_{n\ge 1}$ and ${\left({v}_{n}\right)}_{n\ge 1}$ approximating the Euler-Mascheroni constant. Also, we give the upper and lower bounds for ${\sum }_{k=1}^{n}\frac{1}{k}-\frac{1}{2}ln\left({n}^{2}+n+\frac{1}{3}\right)-\gamma$ 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 $n\in \mathbb{N}:=\left\{1,2,3,\dots \right\}$ by

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

Several bounds for ${D}_{n}-\gamma$ have been given in the literature [17]. For example, the following bounds for ${D}_{n}-\gamma$ were established in [3, 7]:

$\frac{1}{2\left(n+1\right)}<{D}_{n}-\gamma <\frac{1}{2n}\phantom{\rule{1em}{0ex}}\left(n\in \mathbb{N}\right).$

The convergence of the sequence ${D}_{n}$ to γ is very slow. Some quicker approximations to the Euler-Mascheroni constant were established in [821]. For example, Cesàro [8] proved that for every positive integer $n\ge 1$, there exists a number ${c}_{n}\in \left(0,1\right)$ such that the following approximation is valid:

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

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

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

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

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),

$\sum _{k=1}^{n}\frac{1}{k}-ln\left(n+\frac{1}{2}+\frac{1}{24n}-\frac{1}{48{n}^{2}}+\frac{23}{5,760{n}^{3}}\right)=\gamma +O\left({n}^{-5}\right)\phantom{\rule{1em}{0ex}}\left(n\to \mathrm{\infty }\right),$
(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

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

${P}_{n}=\sum _{k=1}^{n}\frac{1}{k}-\frac{1}{2}ln\left({n}^{2}+n+\frac{1}{3}\right)$
(1.7)

and

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

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

$\frac{1}{180{\left(n+1\right)}^{4}}<\gamma -{P}_{n}<\frac{1}{180{n}^{4}}$
(1.8)

and

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

Now we define the sequences

${u}_{n}=\sum _{k=1}^{n}\frac{1}{k}-\frac{1}{2}ln\left({n}^{2}+n+\frac{1}{3}\right)-\frac{1}{r\left({n}^{2}+n+\frac{1}{3}\right)+s{\left({n}^{2}+n+\frac{1}{3}\right)}^{2}+t}$
(1.9)

and

$\begin{array}{rcl}{v}_{n}& =& \sum _{k=1}^{n}\frac{1}{k}-\frac{1}{2}ln\left({n}^{2}+n+\frac{1}{3}\right)\\ -\left(\frac{a}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{2}}+\frac{b}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{3}}+\frac{c}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{4}}+\frac{d}{{\left({n}^{2}+n+\frac{1}{3}\right)}^{5}}\right),\end{array}$
(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 ${\left({u}_{n}\right)}_{n\ge 1}$ and ${\left({v}_{n}\right)}_{n\ge 1}$ approximating the Euler-Mascheroni constant.

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

$r=-\frac{640}{7},\phantom{\rule{2em}{0ex}}s=-180,\phantom{\rule{2em}{0ex}}t=\frac{26,770}{441},$

we have

$\underset{n\to \mathrm{\infty }}{lim}{n}^{11}\left({u}_{n}-{u}_{n+1}\right)=\frac{457,528}{123,773,265}$
(1.11)

and

$\underset{n\to \mathrm{\infty }}{lim}{n}^{10}\left({u}_{n}-\gamma \right)=\frac{457,528}{123,773,265}.$
(1.12)

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

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

$a=-\frac{1}{180},\phantom{\rule{2em}{0ex}}b=\frac{8}{2,835},\phantom{\rule{2em}{0ex}}c=-\frac{5}{1,512},\phantom{\rule{2em}{0ex}}d=\frac{592}{93,555},$

we have

$\underset{n\to \mathrm{\infty }}{lim}{n}^{13}\left({v}_{n}-{v}_{n+1}\right)=-\frac{796,801}{3,648,645}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{n}^{12}\left({v}_{n}-\gamma \right)=-\frac{796,801}{43,783,740}.$

The speed of convergence of the sequence ${\left({v}_{n}\right)}_{n\ge 1}$ 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 $\mathrm{\Gamma }\left(z\right)$ thanks to the Weierstrass formula:

$\mathrm{\Gamma }\left(z\right)=\frac{{e}^{-\gamma z}}{z}\prod _{k=1}^{\mathrm{\infty }}\left\{{\left(1+\frac{z}{k}\right)}^{-1}{e}^{z/k}\right\}\phantom{\rule{1em}{0ex}}\left(z\in \mathbb{C}\setminus {Z}_{0}^{-};{Z}_{0}^{-}:=\left\{-1,-2,-3,\dots \right\}\right).$

The logarithmic derivative of the gamma function

$\psi \left(z\right)=\frac{{\mathrm{\Gamma }}^{\mathrm{\prime }}\left(z\right)}{\mathrm{\Gamma }\left(z\right)}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}ln\mathrm{\Gamma }\left(z\right)={\int }_{1}^{z}\psi \left(t\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t$

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

${\psi }^{\left(n\right)}\left(z\right):=\frac{{\mathrm{d}}^{n}}{\mathrm{d}{z}^{n}}\left\{\psi \left(z\right)\right\}\phantom{\rule{1em}{0ex}}\left(n\in \mathbb{N}\right)$

are called the polygamma functions.

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

$\psi \left(z+1\right)=\psi \left(z\right)+\frac{1}{z}$
(2.1)

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

(2.2)

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

$\psi \left(n+1\right)\sim lnn+\frac{1}{2n}-\frac{1}{12{n}^{2}}+\frac{1}{120{n}^{4}}-\frac{1}{252{n}^{6}}+\cdots \phantom{\rule{1em}{0ex}}\left(n\to \mathrm{\infty }\right).$
(2.3)

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

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

Lemma 1 [24, 25]

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

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

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 $x>0$,

${S}_{k}\left(2n;x\right)<{\left(-1\right)}^{k+1}{\psi }^{\left(k\right)}\left(x\right)<{S}_{k}\left(2n+1;x\right),$
(2.4)

where

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

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 $x>0$,

$\begin{array}{r}\frac{1}{x}+\frac{1}{2{x}^{2}}+\frac{1}{6{x}^{3}}-\frac{1}{30{x}^{5}}+\frac{1}{42{x}^{7}}-\frac{1}{30{x}^{9}}+\frac{5}{66{x}^{11}}-\frac{691}{2,730{x}^{13}}\\ \phantom{\rule{1em}{0ex}}<{\psi }^{\mathrm{\prime }}\left(x\right)<\frac{1}{x}+\frac{1}{2{x}^{2}}+\frac{1}{6{x}^{3}}-\frac{1}{30{x}^{5}}+\frac{1}{42{x}^{7}}-\frac{1}{30{x}^{9}}+\frac{5}{66{x}^{11}}-\frac{691}{2,730{x}^{13}}+\frac{7}{6{x}^{15}},\end{array}$

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 ${n}^{-1}$:

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

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

namely if

$r=-\frac{640}{7},\phantom{\rule{2em}{0ex}}s=-180,\phantom{\rule{2em}{0ex}}t=\frac{26,770}{441}.$

Thus, we have

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

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 ${n}^{-1}$:

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

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

namely if

$a=-\frac{1}{180},\phantom{\rule{2em}{0ex}}b=\frac{8}{2,835},\phantom{\rule{2em}{0ex}}c=-\frac{5}{1,512},\phantom{\rule{2em}{0ex}}d=\frac{592}{93,555}.$

Thus, we have

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

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

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

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

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

where

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

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

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

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

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

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

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

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

where

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

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

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

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

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.

## References

1. Alzer H: Inequalities for the gamma and polygamma functions. Abh. Math. Semin. Univ. Hamb. 1998, 68: 363–372. 10.1007/BF02942573

2. Anderson GD, Barnard RW, Richards KC, Vamanamurthy MK, Vuorinen M: Inequalities for zero-balanced hypergeometric functions. Trans. Am. Math. Soc. 1995, 347: 1713–1723. 10.1090/S0002-9947-1995-1264800-3

3. Rippon PJ: Convergence with pictures. Am. Math. Mon. 1986, 93: 476–478. 10.2307/2323478

4. Tims SR, Tyrrell JA: Approximate evaluation of Euler’s constant. Math. Gaz. 1971, 55: 65–67. 10.2307/3613323

5. Tóth L: Problem E3432. Am. Math. Mon. 1991., 98: Article ID 264

6. Tóth L: Problem E3432 (solution). Am. Math. Mon. 1992, 99: 684–685.

7. Young RM: Euler’s constant. Math. Gaz. 1991, 75: 187–190. 10.2307/3620251

8. Cesàro E: Sur la serie harmonique. Nouvelles Ann. Math. 1885, 4: 295–296.

9. Chen C-P: The best bounds in Vernescu’s inequalities for the Euler’s constant. RGMIA Res. Rep. Coll. 2009., 12: Article ID 11. Available online at http://ajmaa.org/RGMIA/v12n3.php

10. Chen C-P: Inequalities and monotonicity properties for some special functions. J. Math. Inequal. 2009, 3: 79–91.

11. Chen C-P: Inequalities for the Euler-Mascheroni constant. Appl. Math. Lett. 2010, 23: 161–164. 10.1016/j.aml.2009.09.005

12. Chen C-P, Li L: Two accelerated approximations to the Euler-Mascheroni constant. Sci. Magna 2010, 6: 102–110.

13. Chen C-P, Mortici C: New sequence converging towards the Euler-Mascheroni constant. Comput. Math. Appl. 2012, 64: 391–398. 10.1016/j.camwa.2011.03.099

14. Chen C-P, Srivastava HM, Li L, Manyama S: Inequalities and monotonicity properties for the psi (or digamma) function and estimates for the Euler-Mascheroni constant. Integral Transforms Spec. Funct. 2011, 22: 681–693. 10.1080/10652469.2010.538525

15. DeTemple DW: A quicker convergence to Euler’s constant. Am. Math. Mon. 1993, 100: 468–470. 10.2307/2324300

16. Mortici C: On new sequences converging towards the Euler-Mascheroni constant. Comput. Math. Appl. 2010, 59: 2610–2614. 10.1016/j.camwa.2010.01.029

17. Mortici C: Improved convergence towards generalized Euler-Mascheroni constant. Appl. Math. Comput. 2010, 215: 3443–3448. 10.1016/j.amc.2009.10.039

18. Negoi T: A faster convergence to the constant of Euler. Gaz. Mat., Ser. A 1997, 15: 111–113. (in Romanian)

19. Vernescu A: A new accelerated convergence to the constant of Euler. Gaz. Mat., Ser. A 1999, 96(17):273–278. (in Romanian)

20. Villarino M: Ramanujan’s harmonic number expansion into negative powers of a triangular number. J. Inequal. Pure Appl. Math. 2008., 9: Article ID 89. Available online at http://www.emis.de/journals/JIPAM/images/245_07_JIPAM/245_07.pdf

21. Yang S: On an open problem of Chen and Mortici concerning the Euler-Mascheroni constant. J. Math. Anal. Appl. 2012, 396: 689–693. 10.1016/j.jmaa.2012.07.007

22. Berndt B 5. In Ramanujan’s Notebooks. Springer, New York; 1998.

23. Abramowitz M, Stegun IA (Eds): Applied Mathematics Series 55 In Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. 9th edition. National Bureau of Standards, Washington; 1972.

24. Mortici C: New approximations of the gamma function in terms of the digamma function. Appl. Math. Lett. 2010, 23: 97–100. 10.1016/j.aml.2009.08.012

25. Mortici C: Product approximations via asymptotic integration. Am. Math. Mon. 2010, 117: 434–441. 10.4169/000298910X485950

26. Alzer H: On some inequalities for the gamma and psi functions. Math. Comput. 1997, 66: 373–389. 10.1090/S0025-5718-97-00807-7

## Acknowledgements

Dedicated to Professor Hari M Srivastava.

## Author information

Authors

### Corresponding author

Correspondence to Cristinel Mortici.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

CM proposed the sequence ${u}_{n}$. CPC proposed the sequence ${v}_{n}$. CM proposed to solve the problems using Lemma 1, while CPC used Lemma 2 in evaluations. Both authors made the computations and verified their corectedness. The authors read and approved the final manuscript.

## Rights and permissions

Reprints and Permissions

Mortici, C., Chen, CP. On the harmonic number expansion by Ramanujan. J Inequal Appl 2013, 222 (2013). https://doi.org/10.1186/1029-242X-2013-222

• Accepted:

• Published:

• DOI: https://doi.org/10.1186/1029-242X-2013-222

### Keywords

• Euler-Mascheroni constant
• harmonic numbers
• inequality
• psi function
• polygamma functions
• asymptotic expansion