Inequalities and asymptotic expansions related to the generalized Somos quadratic recurrence constant

Abstract

In this paper, we give asymptotic expansions and inequalities related to the generalized Somos quadratic recurrence constant, using its relation with the generalized Euler constant.

1 Introduction

Somos’ quadratic recurrence constant is defined (see [1–3]) by

$$ \begin{aligned}[b] \sigma &=\sqrt{1\sqrt{2\sqrt{3\cdots }}}=\prod _{n=1}^{\infty }n ^{1/2^{n}}=\prod _{k=1}^{\infty } \biggl( 1+\frac{1}{k} \biggr) ^{1/2^{k}}= \exp \Biggl\{ \sum_{k=1}^{\infty } \frac{\ln k}{2^{k}} \Biggr\} \\ & =1.66168794 \ldots\end{aligned} $$

(1.1)

or

$$ \sigma =\exp \biggl\{ - \int_{0}^{1}\frac{1-x}{(2-x)\ln x}\,\mathrm{d}x \biggr\} = \exp \biggl\{ - \int_{0}^{1} \int_{0}^{1}\frac{x}{(2-xy)\ln (xy)} \,\mathrm{d}x \, \mathrm{d}y \biggr\} . $$

(1.2)

The constant σ arises in the study of the asymptotic behavior of the sequence

$$ g_{0}=1,\quad\quad g_{n}=ng_{n-1}^{2}, \quad n \in \mathbb{N}:=\{1,2,3,\ldots \}, $$

(1.3)

with the first few terms

$$ g_{0}=1,\quad\quad g_{1}=1,\quad\quad g_{2}=2, \quad\quad g_{3}=12,\quad\quad g_{4}=576,\quad\quad g_{5}=1{,}658{,}880,\quad\quad \ldots . $$

This sequence behaves as follows (see [4, p. 446] and [3, 5]):

$$\begin{aligned} \begin{aligned}[b] g_{n} &\sim \frac{\sigma^{2^{n}}}{n} \biggl(1+ \frac{2}{n}- \frac{1}{n^{2}}+\frac{4}{n^{3}}-\frac{21}{n^{4}}+ \frac{138}{n^{5}}-\frac{1091}{n ^{6}}+\frac{10{,}088}{n^{7}} -\frac{106{,}918}{n^{8}} \\ & \quad{} +\frac{1{,}279{,}220}{n^{9}}-\frac{17{,}070{,}418}{n^{10}}+ \frac{251{,}560{,}472}{n^{11}}- \frac{4{,}059{,}954{,}946}{n^{12}}+\cdots \biggr)^{-1}. \end{aligned} \end{aligned}$$

(1.4)

The constant σ appears in important problems from pure and applied analysis, and it is the motivation for a large number of research papers (see, for example, [1, 6–16]).

Sondow and Hadjicostas [15] introduced and studied the generalized-Euler-constant function $\gamma (z)$, defined by

$$\begin{aligned} \gamma (z)=\sum_{n=1}^{\infty }z^{n-1} \biggl( \frac{1}{n}-\ln \frac{n+1}{n} \biggr) , \end{aligned}$$

(1.5)

where the series converges when $\vert z \vert \leq 1$. Pilehrood and Pilehrood [13] considered the function $z\gamma (z)$ ($\vert z \vert \leq 1$). The function $\gamma (z)$ generalizes both Euler’s constant $\gamma (1)$ and the alternating Euler constant $\ln \frac{4}{\pi } = \gamma (-1)$ [17, 18].

Sondow and Hadjicostas [15] defined the generalized Somos constant

$$\begin{aligned} \sigma_{t}=\sqrt[t]{1\sqrt[t]{2\sqrt[t]{3\sqrt[t]{4 \cdots }}}}=\prod_{n=1}^{\infty }n^{1/t^{n}}= \biggl( \frac{t}{t-1} \biggr) ^{1/(t-1)} \exp \biggl\{ - \frac{1}{t(t-1)}\gamma \biggl( \frac{1}{t} \biggr) \biggr\} , \quad t>1. \end{aligned}$$

(1.6)

Coffey [19] gave the integral and series representations for $\ln \sigma_{t}$:

$$\begin{aligned} \ln \sigma_{t}= \int_{0}^{\infty } \biggl( \frac{e^{-x}}{t-1}+ \frac{1}{1-te ^{x}} \biggr) \frac{\mathrm{d}x}{x} \end{aligned}$$

(1.7)

and

$$\begin{aligned} \ln \sigma_{t}=\frac{1}{t-1}\sum _{k=1}^{\infty }\frac{(-1)^{k-1}}{k} \operatorname{Li}_{k} \biggl( \frac{1}{t} \biggr) =\frac{1}{t-1}\sum _{k=1}^{ \infty }\frac{1}{k} \biggl[ t \operatorname{Li}_{k} \biggl( \frac{1}{t} \biggr) -1 \biggr] \end{aligned}$$

(1.8)

in terms of the polylogarithm function.

It is known (see [15]) that

$$ \gamma \biggl( \frac{1}{2} \biggr) =2\ln \frac{2}{\sigma }, \quad \text{equivalently},\quad \sigma =2\exp \biggl\{ - \frac{1}{2}\gamma \biggl( \frac{1}{2} \biggr) \biggr\} . $$

(1.9)

Thus, formula (1.5) is closely related to Somos’ quadratic recurrence constant σ.

Define

$$\begin{aligned} \gamma_{n}(z)=\sum_{k=1}^{n}z^{k-1} \biggl( \frac{1}{k}-\ln \frac{k+1}{k} \biggr) , \quad \vert z \vert \leq 1. \end{aligned}$$

Mortici [11] proved that for $n\in \mathbb{N}$,

$$\begin{aligned} \frac{270(n +1)}{2^{n}(270n^{3} + 1530n^{2} + 1065n + 6293)}< \gamma \biggl( \frac{1}{2} \biggr) - \gamma_{n} \biggl( \frac{1}{2} \biggr) < \frac{18}{2^{n}(18n ^{2} +84n - 13)} \end{aligned}$$

(1.10)

and

$$\begin{aligned} &\sum_{k=n+1}^{\infty } \frac{1}{2k^{2}\cdot 3^{k-1}}- \frac{22{,}400(n+1)}{3^{n}(44{,}800n ^{4}+280{,}000n^{3}+435{,}120n^{2}+744{,}380n-2{,}477{,}677)} \\ &\quad < \gamma \biggl( \frac{1}{3} \biggr) -\gamma_{n} \biggl( \frac{1}{3} \biggr) < \sum_{k=n+1}^{\infty } \frac{1}{2k^{2}\cdot 3^{k-1}}-\frac{160}{3^{n}(320n ^{3} + 1680n^{2} + 1428n + 3889)}. \end{aligned}$$

(1.11)

Lu and Song [10] improved Mortici’s results and obtained the inequalities:

$$\begin{aligned} \begin{aligned}[b] \frac{690n^{2} + 3524n + 145}{6(2^{n})(n +1)^{2}(115n^{2} + 894n + 779)}&< \gamma \biggl( \frac{1}{2} \biggr) - \gamma_{n} \biggl( \frac{1}{2} \biggr) \\ &< \frac{48n + 127}{3(2^{n})(16n + 85)(n +1)^{2}}\end{aligned} \end{aligned}$$

(1.12)

and

$$\begin{aligned} &\sum_{k=n+1}^{\infty } \frac{1}{2k^{2}\cdot 3^{k-1}}- \frac{987 840n ^{2} + 8 444 340n + 10 946 779}{40(3^{n})(n +1)^{2}(49 392n^{3} + 582 741n^{2} + 1 769 516n + 1 236 167)} \\ &\quad < \gamma \biggl( \frac{1}{3} \biggr) -\gamma_{n} \biggl( \frac{1}{3} \biggr) < \sum_{k=n+1}^{\infty } \frac{1}{2k^{2}\cdot 3^{k-1}}-\frac{1620n^{2} + 6995n + 1847}{40(3^{n})(81n^{2} + 532n + 451)(n +1)^{3}} \end{aligned}$$

(1.13)

for $n\in \mathbb{N}$.

You and Chen [16] further improved inequalities (1.10)–(1.13). Recently, Chen and Han [7] gave new bounds for $\gamma (1/2)-\gamma_{n}(1/2)$:

$$\begin{aligned} &\frac{1}{2^{n}} \biggl( \frac{1}{(n+1)^{2}}- \frac{8}{3(n+1)^{3}}+ \frac{23}{2(n+1)^{4}}-\frac{332}{5(n+1)^{5}}+\frac{479}{(n+1)^{6}}- \frac{29{,}024}{7(n+1)^{7}} \biggr) \\ &\quad < \gamma \biggl( \frac{1}{2} \biggr) -\gamma_{n} \biggl( \frac{1}{2} \biggr) \\ &\quad < \frac{1}{2^{n}} \biggl( \frac{1}{(n+1)^{2}}- \frac{8}{3(n+1)^{3}}+ \frac{23}{2(n+1)^{4}}-\frac{332}{5(n+1)^{5}}+\frac{479}{(n+1)^{6}} \biggr) \end{aligned}$$

(1.14)

for $n\in \mathbb{N}$, and presented the following asymptotic expansion:

$$\begin{aligned} \begin{aligned}[b] &\gamma \biggl( \frac{1}{2} \biggr) - \gamma_{n} \biggl( \frac{1}{2} \biggr) \\ & \quad \sim \frac{1}{2^{n}(n+1)^{2}} \biggl\{ 1- \frac{8}{3(n+1)}+ \frac{23}{2(n+1)^{2}}-\frac{332}{5(n+1)^{3}}+\frac{479}{(n+1)^{4}}- \cdots \biggr\} \end{aligned} \end{aligned}$$

(1.15)

as $n\to \infty$. Moreover, these authors gave a formula for successively determining the coefficients in (1.15).

Chen and Han [7] pointed out that the lower bound in (1.14) is for $n\geq 24$ sharper than the one in (1.12), and the upper bound in (1.14) is for $n\geq 18$ sharper than the one in (1.12),

For any positive integer $m\geq 2$, in this paper we give the asymptotic expansion of $\gamma ( 1/m ) -\gamma_{n} ( 1/m )$ as $n\to \infty$. Based on the result obtained, we establish the inequality for $\gamma ( 1/4 ) -\gamma_{n} ( 1/4 )$. We also consider the asymptotic expansion for $\gamma ( -1 ) - \gamma_{n} ( -1 )$.

2 Lemmas

Lemma 2.1

As $x\to \infty$,

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) -\sum _{j=2}^{m-1} \frac{(-1)^{j}}{j} \frac{1}{x^{j}}\sim A(x)-\frac{1}{m}A(x+1), $$

(2.1)

where $A(x)$ is defined by

$$ A(x)=\sum_{j=m}^{\infty } \frac{a_{j}}{x^{j}} $$

(2.2)

with the coefficients $a_{j}$ given by the recurrence relation

$$\begin{aligned} a_{j}=\frac{(-1)^{j}}{m-1} \Biggl\{ \frac{m}{j}+ \sum_{k=m}^{j-1}(-1)^{k}a _{k}\binom{j-1}{j-k} \Biggr\} , \quad j\geq m. \end{aligned}$$

(2.3)

Here, and throughout this paper, an empty sum is understood to be zero.

Proof

Using the Maclaurin series of $\ln (1+t)$,

$$ \ln (1+t)=\sum_{j=1}^{\infty } \frac{(-1)^{j-1}}{j}t^{j}, \quad -1< t\leq 1, $$

we obtain

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) -\sum _{j=2}^{m-1} \frac{(-1)^{j}}{j} \frac{1}{x^{j}}=\sum_{j=m}^{\infty } \frac{(-1)^{j}}{j}\frac{1}{x^{j}}. $$

(2.4)

In view of (2.4), we can let

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) -\sum _{j=2}^{m-1} \frac{(-1)^{j}}{j} \frac{1}{x^{j}}\sim \sum_{j=m}^{\infty } \frac{a_{j}}{x ^{j}}-\frac{1}{m}\sum_{j=m}^{\infty } \frac{a_{j}}{(x+1)^{j}}, $$

(2.5)

where $a_{j}$ are real numbers to be determined.

Write (2.5) as

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) -\sum _{j=2}^{m-1} \frac{(-1)^{j}}{j} \frac{1}{x^{j}}\sim \sum_{j=m}^{\infty } \frac{a_{j}}{x ^{j}}-\frac{1}{m}\sum_{j=m}^{\infty } \frac{a_{j}}{x^{j}} \biggl( 1+ \frac{1}{x} \biggr) ^{-j}. $$

(2.6)

Direct computation yields

$$\begin{aligned} \sum_{j=m}^{\infty } \frac{a_{j}}{x^{j}} \biggl( 1+\frac{1}{x} \biggr) ^{-j} &=\sum_{j=m}^{\infty } \frac{a_{j}}{x^{j}}\sum_{k=0}^{\infty } \binom{-j}{k}\frac{1}{x^{k}} \\ &=\sum_{j=m}^{\infty }\frac{a_{j}}{x^{j}}\sum _{k=0}^{\infty }(-1)^{k} \binom{k+j-1}{k}\frac{1}{x^{k}} \\ &=\sum_{j=m}^{\infty }\sum _{k=m}^{j}a_{k}(-1)^{j-k} \binom{j-1}{j-k}\frac{1}{x ^{j}}. \end{aligned}$$

(2.7)

It follows from (2.4), (2.6), and (2.7) that

$$ \sum_{j=m}^{\infty } \frac{(-1)^{j}}{j}\frac{1}{x^{j}}\sim \sum_{j=m} ^{\infty } \Biggl\{ a_{j}-\frac{1}{m}\sum _{k=m}^{j}a_{k}(-1)^{j-k} \binom{j-1}{j-k} \Biggr\} \frac{1}{x^{j}}. $$

(2.8)

Equating coefficients of the term $x^{-j}$ on both sides of (2.8) yields

$$ \frac{(-1)^{j}}{j}=a_{j}-\frac{1}{m}\sum _{k=m}^{j}a_{k}(-1)^{j-k} \binom{j-1}{j-k}, \quad j\geq m. $$

(2.9)

For $j=m$, we obtain $a_{m}=\frac{(-1)^{m}}{m-1}$, and for $j\geq m+1$, we have

$$ \frac{(-1)^{j}}{j}=a_{j}-\frac{1}{m} \Biggl[ \sum _{k=m}^{j-1}a_{k}(-1)^{j-k} \binom{j-1}{j-k}+a_{j} \Biggr] , \quad j\geq m+1. $$

We then obtain the recursive formula

$$\begin{aligned} a_{m}=\frac{(-1)^{m}}{m-1}, \quad\quad a_{j}= \frac{(-1)^{j}m}{(m-1)j}+ \frac{1}{m-1}\sum_{k=m}^{j-1}a_{k}(-1)^{j-k} \binom{j-1}{j-k}, \quad j\geq m+1, \end{aligned}$$

which can be written as (2.3). The proof of Lemma 2.1 is complete. □

Lemma 2.2

Let

$$ a(x)=\frac{1}{3x^{4}}-\frac{32}{45x^{5}}\quad \textit{and}\quad b(x)=\frac{1}{3x ^{4}}-\frac{32}{45x^{5}}+\frac{68}{27x^{6}}. $$

(2.10)

Then, for $x\geq 1$,

$$ a(x)-\frac{1}{4}a(x+1)< \frac{1}{x}-\ln \biggl( 1+ \frac{1}{x} \biggr) -\frac{1}{2x ^{2}}+\frac{1}{3x^{3}}< b(x)- \frac{1}{4}b(x+1). $$

(2.11)

Proof

It is well known that for $-1< t\leq 1$ and $m\in \mathbb{N}$,

$$ \sum_{j=1}^{2m}\frac{(-1)^{j-1}}{j}t^{j}< \ln (1+t)< \sum_{j=1}^{2m-1} \frac{(-1)^{j-1}}{j}t^{j}, $$

which implies that for $x\geq 1$ and $m\geq 2$,

$$ \sum_{j=4}^{2m+1} \frac{(-1)^{j}}{jx^{j}}< \frac{1}{x}-\ln \biggl( 1+ \frac{1}{x} \biggr) - \frac{1}{2x^{2}}+\frac{1}{3x^{3}}< \sum_{j=4}^{2m} \frac{(-1)^{j}}{jx ^{j}}. $$

(2.12)

Using (2.12), we find that

$$\begin{aligned} &\frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) - \frac{1}{2x^{2}}+\frac{1}{3x ^{3}}-a(x)+\frac{1}{4}a(x+1) \\ &\quad >\frac{1}{4x^{4}}-\frac{1}{5x^{5}}-a(x)+\frac{1}{4}a(x+1)= \frac{310x ^{4}+770x^{3}+845x^{2}+445x+92}{180x^{5}(x+1)^{5}}>0 \end{aligned}$$

and

$$\begin{aligned} &\frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) - \frac{1}{2x^{2}}+\frac{1}{3x ^{3}}-b(x)+\frac{1}{4}b(x+1) \\ &\quad < \frac{1}{4x^{4}}-\frac{1}{5x^{5}}+\frac{1}{6x^{6}}-b(x)+ \frac{1}{4}b(x+1) \\ &\quad =-\frac{4380x^{5}+14{,}205x^{4}+21{,}530x^{3}+17{,}439x^{2}+7344x+1270}{540x ^{6}(x+1)^{6}}< 0. \end{aligned}$$

The proof of Lemma 2.2 is complete. □

Remark 2.1

Using the methods from [20–22] it is possible to get estimations (based on the power series expansions) of the logarithm function that can be used, for example, in the analysis of parameterized Euler-constant function, which will be an item for further work.

Lemma 2.3

As $x\to \infty$, we have

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) \sim C(x)+C(x+1), $$

(2.13)

where $C(x)$ is defined by

$$ C(x)=\sum_{j=2}^{\infty } \frac{c_{j}}{x^{j}} $$

(2.14)

with the coefficients $c_{j}$ given by the recurrence relation

$$\begin{aligned} c_{2}=\frac{1}{4}, \quad\quad c_{j}= \frac{(-1)^{j}}{2j}-\frac{1}{2}\sum_{k=2}^{j-1}c_{k}(-1)^{j-k} \binom{j-1}{j-k}, \quad j\geq 3. \end{aligned}$$

(2.15)

Proof

In view of (2.4), we can let

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) \sim \sum_{j=2}^{\infty }\frac{c _{j}}{x^{j}}+\sum _{j=2}^{\infty }\frac{c_{j}}{(x+1)^{j}}, $$

(2.16)

where $c_{j}$ are real numbers to be determined. Write (2.16) as

$$ \frac{1}{x}-\ln \biggl( 1+\frac{1}{x} \biggr) \sim \sum _{j=2}^{\infty }\frac{c _{j}}{x^{j}}+\sum _{j=2}^{\infty }\frac{c_{j}}{x^{j}} \biggl( 1+ \frac{1}{x} \biggr) ^{-j}. $$

Noting that (2.7) holds, we have

$$ \sum_{j=2}^{\infty } \frac{(-1)^{j}}{j}\frac{1}{x^{j}}\sim \sum_{j=2} ^{\infty } \Biggl\{ c_{j}+\sum_{k=2}^{j}c_{k}(-1)^{j-k} \binom{j-1}{j-k} \Biggr\} \frac{1}{x ^{j}}. $$

(2.17)

Equating coefficients of the term $x^{-j}$ on both sides of (2.17) yields

$$ \frac{(-1)^{j}}{j}=c_{j}+\sum_{k=2}^{j}c_{k}(-1)^{j-k} \binom{j-1}{j-k}, \quad j\geq 2. $$

For $j=2$, we obtain $c_{2}=1/4$, and for $j\geq 3$ we have

$$ \frac{(-1)^{j}}{j}=2c_{j}+\sum_{k=2}^{j-1}c_{k}(-1)^{j-k} \binom{j-1}{j-k}, \quad j\geq 3. $$

We then obtain the recursive formula (2.15). The proof of Lemma 2.3 is complete. □

The first few coefficients $c_{j}$ are

$$\begin{aligned}& c_{2}=\frac{1}{4}, \quad\quad c_{3}= \frac{1}{12}, \quad\quad c_{4}=-\frac{1}{8}, \quad\quad c_{5}=- \frac{1}{10},\quad\quad c_{6}= \frac{1}{4}, \\& c_{7}= \frac{17}{56},\quad\quad c_{8}=- \frac{17}{16},\quad\quad c_{9}=-\frac{31}{18}. \end{aligned}$$

3 Main results

For any positive integer $m\geq 2$, Theorem 3.1 gives the asymptotic expansion of $\gamma ( 1/m ) -\gamma_{n} ( 1/m )$ as $n\to \infty$.

Theorem 3.1

For any positive integer $m\geq 2$, we have

$$\begin{aligned} \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n} \biggl( \frac{1}{m} \biggr) \sim \sum_{k=n+1}^{\infty } \sum_{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m ^{k-1}} \frac{1}{k^{j}}+\frac{A(n+1)}{m^{n}},\quad n\to \infty , \end{aligned}$$

(3.1)

where $A(x)$ is given in (2.2). Namely,

$$\begin{aligned} &\gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n} \biggl( \frac{1}{m} \biggr) \\ & \quad \sim \sum_{k=n+1}^{\infty } \sum _{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m ^{k-1}} \frac{1}{k^{j}} \\ &\quad \quad {}+\frac{(-1)^{m}}{m^{n}} \biggl\{ \frac{1}{(m-1)(n+1)^{m}}- \frac{2m ^{2}}{(m+1)(m-1)^{2}(n+1)^{m+1}} \\ &\quad \quad {}+\frac{m^{2}(m^{2}+8m+3)}{2(m+2)(m-1)^{3}(n+1)^{m+2}}-\frac{(m+1)(m ^{3}+12m^{2}+51m+8)m^{2}}{6(m-1)^{4}(m+3)(n+1)^{m+3}} \\ & \quad \quad {}+ \frac{m^{2}(m^{6}+25m^{5}+216m^{4}+866m^{3}+1241m^{2}+501m+30)}{24(m-1)^{5}(m+4)(n+1)^{m+4}}- \cdots \biggr\} . \end{aligned}$$

(3.2)

Proof

Write (2.1) as

$$ \frac{1}{k}-\ln \biggl( 1+\frac{1}{k} \biggr) -\sum _{j=2}^{m-1} \frac{(-1)^{j}}{j} \frac{1}{k^{j}}= A_{N}(k)-\frac{1}{m}A_{N}(k+1)+O \biggl( \frac{1}{k^{N+1}} \biggr) , $$

(3.3)

where

$$ A_{N}(k)=\sum_{j=m}^{N} \frac{a_{j}}{k^{j}} $$

(3.4)

with the coefficients $a_{j}$ given by the recurrence relation (2.3). From (3.3), we have

$$\begin{aligned} &\frac{1}{m^{k-1}} \biggl( \frac{1}{k}-\ln \biggl( 1+ \frac{1}{k} \biggr) \biggr) -\frac{1}{m ^{k-1}}\sum _{j=2}^{m-1}\frac{(-1)^{j}}{j}\frac{1}{k^{j}} \\ & \quad = \frac{A_{N}(k)}{m^{k-1}}-\frac{A_{N}(k+1)}{m^{k}}+O \biggl( \frac{1}{m ^{k-1}k^{N+1}} \biggr) . \end{aligned}$$

(3.5)

Adding (3.5) from $k=n+1$ to $k=\infty$, we have

$$\begin{aligned} &\gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n} \biggl( \frac{1}{m} \biggr) - \sum_{k=n+1}^{\infty } \sum_{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m^{k-1}} \frac{1}{k ^{j}}=\frac{1}{m^{n}} \biggl\{ A_{N}(n+1)+O \biggl( \frac{1}{(n+1)^{N+1}} \biggr) \biggr\} , \end{aligned}$$

which can be written as (3.1). The proof of Theorem 3.1 is complete. □

Remark 3.1

For $m=2$ in (3.2), we obtain (1.15). For $m=3$ in (3.2), we find

$$\begin{aligned} &\gamma \biggl( \frac{1}{3} \biggr) -\gamma_{n} \biggl( \frac{1}{3} \biggr) \\ &\quad \sim \sum_{k=n+1}^{\infty } \frac{1}{2k^{2} 3^{k-1}} \\ &\quad\quad {}+\frac{1}{3^{n}(n+1)^{3}} \biggl\{ -\frac{1}{2}+ \frac{9}{8(n+1)}- \frac{81}{20(n+1)^{2}}+\frac{37}{2(n+1)^{3}}- \frac{5661}{56(n+1)^{4}}+ \cdots \biggr\} . \end{aligned}$$

(3.6)

For $m=4$ in (3.2), we find

$$\begin{aligned} &\gamma \biggl( \frac{1}{4} \biggr) -\gamma_{n} \biggl( \frac{1}{4} \biggr) \\ &\quad \sim \sum_{k=n+1}^{\infty } \biggl( \frac{1}{2}-\frac{1}{3k} \biggr) \frac{1}{k ^{2}4^{k-1}} \\ &\quad\quad {}+\frac{1}{4^{n}(n+1)^{4}} \biggl\{ \frac{1}{3}- \frac{32}{45(n+1)}+ \frac{68}{27(n+1)^{2}}-\frac{2080}{189(n+1)^{3}} \\ &\quad\quad {}+ \frac{9017}{162(n+1)^{4}}-\cdots \biggr\} . \end{aligned}$$

(3.7)

Formula (3.7) motivated us to establish Theorem 3.2.

Theorem 3.2

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

$$\begin{aligned} &\sum_{k=n+1}^{\infty } \biggl( \frac{1}{2}- \frac{1}{3k} \biggr) \frac{1}{k ^{2}4^{k-1}}+ \frac{1}{4^{n}} \biggl\{ \frac{1}{3(n+1)^{4}}- \frac{32}{45(n+1)^{5}} \biggr\} \\ &\quad< \gamma \biggl( \frac{1}{4} \biggr) -\gamma _{n} \biggl( \frac{1}{4} \biggr) \\ &\quad < \sum_{k=n+1}^{\infty } \biggl( \frac{1}{2}-\frac{1}{3k} \biggr) \frac{1}{k ^{2}4^{k-1}}+ \frac{1}{4^{n}} \biggl\{ \frac{1}{3(n+1)^{4}}- \frac{32}{45(n+1)^{5}}+ \frac{68}{27(n+1)^{6}} \biggr\} . \end{aligned}$$

(3.8)

Proof

From the double inequality (2.11), we have

$$ \begin{aligned}[b] \frac{a(k)}{4^{k-1}}-\frac{a(k+1)}{4^{k}}&< \frac{1}{4^{k-1}} \biggl( \frac{1}{k}- \ln \biggl( 1+\frac{1}{k} \biggr) \biggr) - \frac{1}{4^{k-1}} \biggl( \frac{1}{2k ^{2}}- \frac{1}{3k^{3}} \biggr) \\ &< \frac{b(k)}{4^{k-1}}- \frac{b(k+1)}{4^{k}},\end{aligned} $$

(3.9)

where $a(x)$ and $b(x)$ are given in (2.10). Adding inequalities (3.9) from $k=n+1$ to $k=\infty$, we have

$$ \frac{a(n+1)}{4^{n}}< \sum_{k=n+1}^{\infty } \frac{1}{4^{k-1}} \biggl( \frac{1}{k}- \ln \biggl( 1+\frac{1}{k} \biggr) \biggr) -\sum_{k=n+1}^{\infty } \frac{1}{4^{k-1}} \biggl( \frac{1}{2k^{2}}-\frac{1}{3k^{3}} \biggr) < \frac{b(n+1)}{4^{n}}, $$

which can be written as (3.8). The proof of Theorem 3.2 is complete. □

Remark 3.2

Inequality (3.8) can be further refined by inserting additional terms on both sides of the inequality. For example, for $n\in \mathbb{N}$, we have

$$\begin{aligned} &\sum_{k=n+1}^{\infty } \biggl( \frac{1}{2}- \frac{1}{3k} \biggr) \frac{1}{k ^{2}4^{k-1}}+ \frac{1}{4^{n}} \biggl\{ \frac{1}{3(n+1)^{4}}- \frac{32}{45(n+1)^{5}}+ \frac{68}{27(n+1)^{6}}- \frac{2080}{189(n+1)^{7}} \biggr\} \\ &\quad < \gamma \biggl( \frac{1}{4} \biggr) -\gamma_{n} \biggl( \frac{1}{4} \biggr) \\ &\quad < \sum_{k=n+1}^{\infty } \biggl( \frac{1}{2}-\frac{1}{3k} \biggr) \frac{1}{k ^{2}4^{k-1}} \\ & \qquad {}+\frac{1}{4^{n}} \biggl\{ \frac{1}{3(n+1)^{4}}-\frac{32}{45(n+1)^{5}}+ \frac{68}{27(n+1)^{6}}-\frac{2080}{189(n+1)^{7}}+ \frac{9017}{162(n+1)^{8}} \biggr\} . \end{aligned}$$

(3.10)

Remark 3.3

Following the same method as the one used in the proof of Theorem 3.2, we can prove the following inequality:

$$\begin{aligned} &\sum_{k=n+1}^{\infty } \frac{1}{2k^{2}3^{k-1}}+ \frac{1}{3^{n}} \biggl\{ -\frac{1}{2(n+1)^{3}}+ \frac{9}{8(n+1)^{4}}- \frac{81}{20(n+1)^{5}}+ \frac{37}{2(n+1)^{6}}-\frac{5661}{56(n+1)^{7}} \biggr\} \\ &\quad < \gamma \biggl( \frac{1}{3} \biggr) -\gamma_{n} \biggl( \frac{1}{3} \biggr) \\ &\quad < \sum_{k=n+1}^{\infty }\frac{1}{2k^{2} 3^{k-1}}+ \frac{1}{3^{n}} \biggl\{ -\frac{1}{2(n+1)^{3}}+\frac{9}{8(n+1)^{4}}- \frac{81}{20(n+1)^{5}}+\frac{37}{2(n+1)^{6}} \biggr\} \end{aligned}$$

(3.11)

for $n\in \mathbb{N}$. We omit the proof.

In view of (1.14), (3.11), (3.8), and (3.10), we pose the following conjecture.

Conjecture 3.1

For any positive integer $m\geq 2$, we have

$$\begin{aligned} \begin{aligned}[b] \frac{(-1)^{m}}{m^{n}}\sum_{j=m}^{2N} \frac{a_{j}}{(n+1)^{j}} &< (-1)^{m} \Biggl\{ \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n} \biggl( \frac{1}{m} \biggr) - \sum_{k=n+1}^{\infty }\sum _{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m^{k-1}k ^{j}} \Biggr\} \\ &< \frac{(-1)^{m}}{m^{n}}\sum_{j=m}^{2N+1} \frac{a_{j}}{(n+1)^{j}}, \end{aligned} \end{aligned}$$

(3.12)

with the coefficients $a_{j}$ given in (2.3).

By using the Maple software, we find, as $n\to \infty$,

$$\begin{aligned}& \gamma \biggl( \frac{1}{2} \biggr) -\gamma_{n} \biggl( \frac{1}{2} \biggr) \sim \frac{1}{2^{n}(n+1)^{2}} \biggl(1+ \frac{-\frac{8}{3}}{n+ \frac{85}{16}}+\frac{-\frac{2689}{160}}{ (n+\frac{807{,}797}{129{,}072} )^{3}}+\cdots \biggr), \end{aligned}$$

(3.13)

$$\begin{aligned}& \begin{aligned}[b] & \gamma \biggl( \frac{1}{3} \biggr) - \gamma_{n} \biggl( \frac{1}{3} \biggr) \\ & \quad \sim \sum_{k=n+1}^{\infty } \frac{1}{2k^{2} 3^{k-1}}+ \frac{1}{3^{n}(n+1)^{3}} \biggl(-\frac{1}{2}+ \frac{\frac{9}{8}}{n+ \frac{23}{5}}+ \frac{\frac{98}{25}}{ (n+\frac{140{,}843}{27{,}440} )^{3}}+\cdots \biggr) \end{aligned} \end{aligned}$$

(3.14)

and

$$\begin{aligned} \gamma \biggl( \frac{1}{4} \biggr) -\gamma_{n} \biggl( \frac{1}{4} \biggr) & \sim \sum_{k=n+1}^{\infty } \biggl( \frac{1}{2}-\frac{1}{3k} \biggr) \frac{1}{k ^{2}4^{k-1}} \\ &\quad {}+\frac{1}{4^{n}(n+1)^{4}} \biggl\{ \frac{1}{3}+\frac{- \frac{32}{45}}{n+\frac{109}{24}}+ \frac{-\frac{2365}{1134}}{ (n+ \frac{825{,}361}{170{,}280} )^{3}}+\cdots \biggr\} . \end{aligned}$$

(3.15)

From a computational viewpoint, formulas (3.13), (3.14), and (3.15) improve formulas (1.15), (3.6), and (3.7), respectively.

For any positive integer $m\geq 2$, we here provide a pair of recurrence relations for determining the constants $p_{\ell }\equiv p_{\ell }(m)$ and $q_{\ell }\equiv q_{\ell }(m)$ (see Remark 3.4) such that

$$\begin{aligned} \begin{aligned}[b]& \gamma \biggl( \frac{1}{m} \biggr) - \gamma_{n} \biggl( \frac{1}{m} \biggr) \\ & \quad \sim \sum_{k=n+1}^{\infty } \sum _{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m ^{k-1}} \frac{1}{k^{j}}+ \frac{1}{m^{n}(n+1)^{m}} \Biggl( a_{m}+\sum_{ \ell =1}^{\infty } \frac{p_{\ell }}{(n+q_{\ell })^{2\ell -1}} \Biggr) \end{aligned} \end{aligned}$$

(3.16)

as $n\to \infty$. This develops formulas (3.13), (3.14), and (3.15) to produce a general result given by Theorem 3.3.

Theorem 3.3

For any positive integer $m\geq 2$, we have

$$\begin{aligned} \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n-1} \biggl( \frac{1}{m} \biggr) \sim \sum_{k=n}^{\infty } \sum_{j=2}^{m-1} \frac{(-1)^{j}}{j\cdot m^{k-1}} \frac{1}{k^{j}}+\frac{1}{m^{n-1}n^{m}} \Biggl( a_{m}+\sum _{\ell =1}^{\infty }\frac{\lambda_{\ell }}{(n+\mu_{ \ell })^{2\ell -1}} \Biggr) \end{aligned}$$

(3.17)

as $n\to \infty$, where $\lambda_{\ell }\equiv \lambda_{\ell }(m)$ and $\mu_{\ell }\equiv \mu_{\ell }(m)$ are given by a pair of recurrence relations

$$ \lambda_{\ell }=a_{m+2\ell -1}-\sum _{k=1}^{\ell -1}\lambda_{k}\mu_{k} ^{2\ell -2k}\binom{2\ell -2}{2\ell -2k}, \quad \ell \geq 2, $$

(3.18)

and

$$\begin{aligned} \mu_{\ell }=-\frac{1}{(2\ell -1)\lambda_{\ell }} \Biggl\{ a_{m+2\ell }+ \sum_{k=1}^{\ell -1} \lambda_{k}\mu_{k}^{2\ell -2k+1}\binom{2\ell -1}{2 \ell -2k+1} \Biggr\} , \quad \ell \geq 2, \end{aligned}$$

(3.19)

with

$$\begin{aligned} \lambda_{1}=a_{m+1}=\frac{(-1)^{m+1}2m^{2}}{(m+1)(m-1)^{2}}\quad \textit{and}\quad \mu_{1}=-\frac{a_{m+2}}{\lambda_{1}}=\frac{(m+1)(m ^{2}+8m+3)}{4(m+2)(m-1)}. \end{aligned}$$

Here $a_{j}$ are given in (2.3).

Proof

In view of (3.13), (3.14), and (3.15), we let

$$\begin{aligned} \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n-1} \biggl( \frac{1}{m} \biggr) \sim \sum_{k=n}^{\infty } \sum_{j=2}^{m-1} \frac{(-1)^{j}}{j\cdot m^{k-1}} \frac{1}{k^{j}}+\frac{1}{m^{n-1}n^{m}} \Biggl( a_{m}+\sum _{\ell =1}^{\infty }\frac{\lambda_{\ell }}{(n+\mu_{ \ell })^{2\ell -1}} \Biggr) , \end{aligned}$$

where $\lambda_{\ell }$ and $\mu_{\ell }$ are real numbers to be determined. This can be written as

$$\begin{aligned} \begin{aligned}[b] & m^{n-1}n^{m} \Biggl\{ \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n-1} \biggl( \frac{1}{m} \biggr) - \sum_{k=n}^{\infty } \sum_{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m^{k-1}} \frac{1}{k ^{j}} \Biggr\} \\ & \quad \sim a_{m}+\sum_{j=1}^{\infty } \frac{\lambda_{j}}{n^{2j-1}} \biggl( 1+\frac{\mu_{j}}{n} \biggr) ^{-2j+1}. \end{aligned} \end{aligned}$$

(3.20)

Direct computation yields

$$\begin{aligned} \sum_{j=1}^{\infty }\frac{\lambda_{j}}{n^{2j-1}} \biggl( 1+ \frac{\mu_{j}}{n} \biggr) ^{-2j+1} &=\sum _{j=1}^{\infty }\frac{\lambda _{j}}{n^{2j-1}}\sum _{k=0}^{\infty }\binom{-2j+1}{k}\frac{\mu_{j}^{k}}{n ^{k}} \\ &= \sum_{j=1}^{\infty }\frac{\lambda_{j}}{n^{2j-1}}\sum _{k=0}^{ \infty }(-1)^{k} \binom{k+2j-2}{k}\frac{\mu_{j}^{k}}{n^{k}} \\ &= \sum_{j=1}^{\infty }\sum _{k=0}^{j-1}\lambda_{k+1}\mu_{k+1}^{j-k-1}(-1)^{j-k-1} \binom{j+k-1}{j-k-1}\frac{1}{n^{j+k}}, \end{aligned}$$

which can be written as

$$\begin{aligned} \sum_{j=1}^{\infty } \frac{\lambda_{j}}{n^{2j-1}} \biggl( 1+ \frac{\mu_{j}}{n} \biggr) ^{-2j+1} \sim \sum_{j=1}^{\infty } \Biggl\{ \sum _{k=1} ^{\lfloor \frac{j+2}{2}\rfloor }\lambda_{k}\mu_{k}^{j-2k+1}(-1)^{j-1} \binom{j-1}{j-2k+1} \Biggr\} \frac{1}{n^{j}}. \end{aligned}$$

(3.21)

Substituting (3.21) into (3.20) we have

$$\begin{aligned} &m^{n-1}n^{m} \Biggl\{ \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n-1} \biggl( \frac{1}{m} \biggr) - \sum _{k=n}^{\infty }\sum_{j=2}^{m-1} \frac{(-1)^{j}}{j\cdot m^{k-1}}\frac{1}{k ^{j}} \Biggr\} \\ & \quad \sim a_{m}+\sum_{j=1}^{\infty } \Biggl\{ \sum_{k=1}^{\lfloor \frac{j+2}{2}\rfloor } \lambda_{k}\mu_{k}^{j-2k+1}(-1)^{j-1} \binom{j-1}{j-2k+1} \Biggr\} \frac{1}{n^{j}}. \end{aligned}$$

(3.22)

On the other hand, it follows from (3.1) that

$$\begin{aligned} m^{n-1}n^{m} \Biggl\{ \gamma \biggl( \frac{1}{m} \biggr) -\gamma_{n-1} \biggl( \frac{1}{m} \biggr) - \sum_{k=n}^{\infty }\sum _{j=2}^{m-1}\frac{(-1)^{j}}{j\cdot m^{k-1}}\frac{1}{k ^{j}} \Biggr\} \sim \sum_{j=0}^{\infty } \frac{a_{m+j}}{n^{j}}. \end{aligned}$$

(3.23)

Equating coefficients of the term $n^{-j}$ on the right-hand sides of (3.22) and (3.23), we obtain

$$ a_{m+j}=\sum_{k=1}^{\lfloor \frac{j+2}{2}\rfloor } \lambda_{k}\mu_{k} ^{j-2k+1}(-1)^{j-1} \binom{j-1}{j-2k+1}, \quad j\in \mathbb{N}. $$

(3.24)

Setting $j=2\ell -1$ and $j=2\ell$ in (3.24), respectively, yields

$$ a_{m+2\ell -1}=\sum_{k=1}^{\ell } \lambda_{k}\mu_{k}^{2\ell -2k}\binom{2 \ell -2}{2\ell -2k} $$

(3.25)

and

$$\begin{aligned} a_{m+2\ell } &=-\sum_{k=1}^{\ell +1} \lambda_{k}\mu_{k}^{2\ell -2k+1}\binom{2 \ell -1}{2\ell -2k+1} \\ &=-\sum_{k=1}^{\ell }\lambda_{k} \mu_{k}^{2\ell -2k+1}\binom{2\ell -1}{2 \ell -2k+1}-\lambda_{\ell +1} \mu_{\ell +1}^{-1}\binom{2\ell -1}{-1} \\ &=-\sum_{k=1}^{\ell }\lambda_{k} \mu_{k}^{2\ell -2k+1}\binom{2\ell -1}{2 \ell -2k+1}. \end{aligned}$$

(3.26)

For $\ell =1$, from (3.25) and (3.26) we obtain

$$\begin{aligned} \lambda_{1}=a_{m+1}=\frac{(-1)^{m+1}2m^{2}}{(m+1)(m-1)^{2}}\quad \text{and}\quad \mu_{1}=-\frac{a_{m+2}}{\lambda_{1}}=\frac{(m+1)(m ^{2}+8m+3)}{4(m+2)(m-1)}, \end{aligned}$$

and for $\ell \geq 2$ we have

$$ a_{m+2\ell -1}=\sum_{k=1}^{\ell -1} \lambda_{k}\mu_{k}^{2\ell -2k}\binom{2 \ell -2}{2\ell -2k}+ \lambda_{\ell } $$

and

$$\begin{aligned} a_{m+2\ell } &=-\sum_{k=1}^{\ell -1} \lambda_{k}\mu_{k}^{2\ell -2k+1}\binom{2 \ell -1}{2\ell -2k+1}-(2\ell -1) \lambda_{\ell }\mu_{\ell }. \end{aligned}$$

We then obtain the recurrence relations (3.18) and (3.19). The proof of Theorem 3.3 is complete. □

Here we give explicit numerical values of some first terms of $\lambda_{\ell }$ and $\mu_{\ell }$ by using formulas (3.18) and (3.19). This shows how easily we can determine the constants $a_{\ell }$ and $b_{\ell }$ in (3.17).

$$\begin{aligned}& \lambda_{1}=a_{m+1}=\frac{(-1)^{m+1}2m^{2}}{(m+1)(m-1)^{2}}, \\& \mu_{1}=-\frac{a_{m+2}}{\lambda_{1}}= \frac{(m+1)(m^{2}+8m+3)}{4(m+2)(m-1)}, \\& \begin{aligned} \lambda_{2} &=a_{m+3}-\lambda_{1} \mu_{1}^{2} \\ &=\frac{(-1)^{m+1}m^{2}(m+1)(m^{3}+12m^{2}+51m+8)}{6(m-1)^{4}(m+3)} \\ &\quad {}-\frac{(-1)^{m+1}2m^{2}}{(m+1)(m-1)^{2}} \biggl( \frac{(m+1)(m ^{2}+8m+3)}{4(m+2)(m-1)} \biggr) ^{2} \\ &=(-1)^{m+1} \frac{m^{2}(m+1)(m^{5}+7m^{4}+58m^{3}+266m^{2}+485m+47)}{24(m-1)^{4}(m+3)(m+2)^{2}}, \end{aligned} \\& \begin{aligned} \mu_{2} &=-\frac{a_{m+4}+\lambda_{1}\mu_{1}^{3}}{3\lambda_{2}} \\ &=-\frac{\frac{(-1)^{m}m^{2}(m^{6}+25m^{5}+216m^{4}+866m^{3}+1241m ^{2}+501m+30)}{24(m-1)^{5}(m+4)}+\lambda_{1}\mu_{1}^{3}}{3\lambda_{2}} \\ &=\bigl((m+3) \bigl(m^{9}+34m^{8}+450m^{7}+3634m^{6}+17{,}584m^{5}+48{,}642m ^{4}+71{,}302m^{3} \\ &\quad{} +50{,}926m^{2}+14{,}151m+636\bigr)\bigr) /\bigl(12(m+2) (m+4) \bigl(m^{5}+7m^{4}+58m ^{3}+266m^{2} \\ &\quad{}+485m+47\bigr) \bigl(m^{2}-1\bigr)\bigr). \end{aligned} \end{aligned}$$

Remark 3.4

The constants $p_{\ell }$ and $q_{\ell }$ in (3.16) are given by

$$ p_{\ell }:=\lambda_{\ell }\quad \text{and}\quad q_{\ell }:=1+\mu_{ \ell }. $$

Setting $m=2, 3$, and 4 in (3.16), respectively, yields (3.13), (3.14), and (3.15).

Noting that $\ln \frac{4}{\pi } = \gamma (-1)$ holds, Theorem 3.4 presents the asymptotic expansion for $\ln \frac{4}{\pi }$.

Theorem 3.4

As $n\to \infty$, we have

$$ \gamma ( -1 ) -\gamma_{n} ( -1 ) \sim (-1)^{n}C(n+1), $$

(3.27)

where $C(x)$ is given in (2.14). Namely,

$$\begin{aligned} \begin{aligned}[b] &\gamma ( -1 ) -\gamma_{n} ( -1 ) \\ &\quad \sim (-1)^{n} \biggl\{ \frac{1}{4(n+1)^{2}}+\frac{1}{12(n+1)^{3}}- \frac{1}{8(n+1)^{4}}- \frac{1}{10(n+1)^{5}}+\cdots \biggr\} .\end{aligned} \end{aligned}$$

(3.28)

Proof

Write (2.13) as

$$ \frac{1}{k}-\ln \biggl( 1+\frac{1}{k} \biggr) =C_{N}(k)+C_{N}(k+1)+O \biggl( \frac{1}{k ^{N+1}} \biggr) , $$

(3.29)

where

$$ C_{N}(x)=\sum_{j=2}^{N} \frac{c_{j}}{x^{j}} $$

(3.30)

with the coefficients $c_{j}$ given by the recurrence relation (2.15).

From (3.29), we have

$$ (-1)^{k-1} \biggl( \frac{1}{k}-\ln \biggl( 1+ \frac{1}{k} \biggr) \biggr) =(-1)^{k-1}C _{N}(k)-(-1)^{k}C_{N}(k+1)+O \biggl( \frac{(-1)^{k-1}}{k^{N+1}} \biggr) . $$

(3.31)

Adding (3.31) from $k=n+1$ to $k=\infty$, we have

$$\begin{aligned} \gamma ( -1 ) -\gamma_{n} ( -1 ) &=\sum_{k=n+1}^{ \infty }(-1)^{k-1} \biggl( \frac{1}{k}-\ln \biggl( 1+\frac{1}{k} \biggr) \biggr) \\ &=(-1)^{n}C_{N}(n+1)+O \biggl( \frac{1}{(n+1)^{N+1}} \biggr) , \end{aligned}$$

(3.32)

which can be written as (3.27). The proof of Theorem 3.4 is complete. □

Remark 3.5

We see from (3.28) that the alternating Euler constant $\ln \frac{4}{\pi }$ has the following expansion:

$$\begin{aligned} \ln \frac{4}{\pi } &\sim \sum_{k=1}^{n}(-1)^{k-1} \biggl( \frac{1}{k}- \ln \frac{k+1}{k} \biggr) \\ &\quad {}+(-1)^{n} \biggl\{ \frac{1}{4(n+1)^{2}}+\frac{1}{12(n+1)^{3}}- \frac{1}{8(n+1)^{4}}-\frac{1}{10(n+1)^{5}}+\cdots \biggr\} . \end{aligned}$$

(3.33)

4 Conclusions

In this paper, we give asymptotic expansions related to the generalized Somos quadratic recurrence constant (Theorems 3.1 and 3.3). We present the inequalities for $\gamma ( \frac{1}{4} ) - \gamma_{n} ( \frac{1}{4} )$ and $\gamma ( \frac{1}{3} ) - \gamma_{n} ( \frac{1}{3} )$ (see (3.8), (3.10), and (3.11)). The expansion of the alternating Euler constant $\ln \frac{4}{\pi }$ is also obtained (see (3.33)).