Some sharp continued fraction inequalities for the Euler-Mascheroni constant

Abstract

The aim of this paper is to establish some new continued fraction inequalities for the Euler-Mascheroni constant by multiple-correction method.

1 Introduction

Euler introduced a constant, then later this constant was called ‘Euler’s constant’ as the limit of the sequence

$$ \gamma(n):=\sum_{m=1}^{n} \frac{1}{m} -\ln n. $$

(1.1)

It is also known as the Euler-Mascheroni constant. There are many famous unsolved problems about the nature of this constant (see e.g. Dence and Dence [1], Havil [2] and Lagarias [3]). For example, it is a long-standing open problem if it is a rational number. A good part of its mystery comes from the fact that the known algorithms converging to γ are not very fast, at least, when they are compared to similar algorithms for π and e.

The sequence $(\gamma(n) )_{n\in\mathbb {N}}$ converges very slowly toward γ, like $(2n)^{-1}$, by Young (see [4]). Up to now, many authors are preoccupied to improve its rate of convergence, see e.g. [1, 4–13] and the references therein. We list some main results as follows:

$$\begin{aligned}& \sum_{m=1}^{n}\frac{1}{m} -\ln \biggl(n+\frac{1}{2} \biggr)=\gamma+O\bigl(n^{-2}\bigr)\quad \mbox{(DeTemple [9])}, \\& \sum_{m=1}^{n}\frac{1}{m} -\ln \frac{n^{3}+\frac{3}{2}n^{2}+\frac{227}{240}+\frac{107}{480}}{ n^{2}+n+\frac{97}{240}}=\gamma+O\bigl(n^{-6}\bigr)\quad \mbox{(Mortici [4])}, \\& \sum_{m=1}^{n}\frac{1}{m} -\ln \biggl( 1+\frac{1}{2n}+\frac{1}{24n^{2}}-\frac{1}{48n^{3}} +\frac{23}{5\text{,}760n^{4}} \biggr) \\& \quad =\gamma+O\bigl(n^{-5}\bigr)\quad \mbox{(Chen and Mortici [6])}. \end{aligned}$$

Recently, Mortici and Chen [5] provided a very interesting sequence

$$\begin{aligned} \nu(n) =&\sum_{m=1}^{n}\frac{1}{m} - \frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &{}- \biggl(\frac{-\frac{1}{180}}{ (n^{2}+n+\frac{1}{3} )^{2}} +\frac{\frac{8}{2\text{,}835}}{ (n^{2}+n+\frac{1}{3} )^{3}}+ +\frac{\frac{5}{1\text{,}512}}{ (n^{2}+n+\frac{1}{3} )^{4}} + \frac{\frac{592}{93\text{,}555}}{ (n^{2}+n+\frac{1}{3} )^{5}} \biggr) \end{aligned}$$

and proved

$$ \lim_{n\rightarrow\infty}n^{12} \bigl(\nu(n)-\gamma \bigr)=-\frac {796\text{,}801}{43\text{,}783\text{,}740}. $$

(1.2)

Hence the rate of convergence of the sequence $(\nu(n) )_{n\in\mathbb{N}}$ is $n^{-12}$.

Very recently, by inserting the continued fraction term in (1.1), Lu [11] introduced a class of sequences $(r_{k}(n) )_{n\in\mathbb{N}}$ and showed

$$\begin{aligned}& \frac{1}{72(n+1)^{3}}< \gamma-r_{2}(n)< \frac{1}{72n^{3}}, \end{aligned}$$

(1.3)

$$\begin{aligned}& \frac{1}{120(n+1)^{4}}< r_{3}(n)-\gamma< \frac{1}{120(n-1)^{4}}. \end{aligned}$$

(1.4)

It is their works that motivate our study. In this paper, starting from the sequence $\nu(n)$, based one the works of Mortici, Chen and Lu, we provide some new classes of convergent sequences with faster rate of convergence for the Euler-Mascheroni constant as follows.

Theorem 1

For the Euler-Mascheroni constant, we have the following convergent sequence:

$$\begin{aligned} r_{k}(n) =&\sum_{m=1}^{n} \frac{1}{m}-\frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &{}-\frac{a_{1}}{(n+\frac{1}{2})^{4}+b_{2} (n+\frac{1}{2})^{2}+b_{0}+}\frac {a_{2}}{(n+\frac{1}{2})^{2}+k_{2}+}\ddots, \end{aligned}$$

where

$$\begin{aligned}& b_{2}=\frac{85}{126}, \qquad b_{0}=-\frac{18\text{,}287}{63\text{,}504}; \\& a_{1}=-\frac{1}{180}, \qquad a_{2}=\frac{1\text{,}830\text{,}112}{2\text{,}750\text{,}517}, \qquad a_{3}=-\frac {36\text{,}637\text{,}398\text{,}233\text{,}630\text{,}775}{9\text{,}056\text{,}534\text{,}057\text{,}598\text{,}976}, \\& a_{4}=-\frac {16\text{,}486\text{,}938\text{,}208\text{,}076\text{,}386\text{,}182\text{,}240\text{,}455\text{,}433\text{,}101\text{,}197\text{,}312}{1\text{,}114\text{,}333\text{,}428\text{,}433\text{,}648\text{,}110\text{,}295\text{,}680\text{,}727\text{,}490\text{,}570\text{,}625}, \\& a_{5}=-\frac{202\ldots665}{538\ldots376};\qquad \ldots, \\& k_{2}=\frac{19\text{,}949\text{,}142\text{,}781}{5\text{,}995\text{,}446\text{,}912},\qquad k_{3}=\frac {83\text{,}116\text{,}192\text{,}006\text{,}963\text{,}596\text{,}639\text{,}745\text{,}097}{13\text{,}306\text{,}111\text{,}671\text{,}966\text{,}702\text{,}431\text{,}356\text{,}800}, \\& k_{4}=\frac {3\text{,}223\text{,}193\text{,}895\text{,}482\text{,}188\text{,}285\text{,}536\text{,}076\text{,}617\text{,}166\text{,}535\text{,}131\text{,}877\text{,}815\text{,}854\text{,}443}{314\text{,}899\text{,}884\text{,}866\text{,}916\text{,}635\text{,}406\text{,}301\text{,}198\text{,}738\text{,}123\text{,}952\text{,}960\text{,}626\text{,}300\text{,}800}, \\& k_{5}=\frac{155\ldots 06831}{102\ldots440},\qquad \ldots. \end{aligned}$$

For $1\le l\le5$, we have

$$\begin{aligned}& \lim_{n\rightarrow\infty}n^{11} \bigl(r_{1}(n)- \gamma \bigr)=\frac {457\text{,}528}{123\text{,}773\text{,}265}:=C_{1}, \\& \lim_{n\rightarrow\infty}n^{15} \bigl(r_{2}(n)-\gamma \bigr)=\frac {16\text{,}615\text{,}600\text{,}105\text{,}955}{1\text{,}111\text{,}124\text{,}185\text{,}307\text{,}136}:=C_{2}, \\& \lim_{n\rightarrow\infty}n^{19} \bigl(r_{3}(n)-\gamma \bigr)=\frac {10\text{,}827\text{,}769\text{,}830\text{,}530\text{,}486\text{,}830\text{,}966\text{,}003\text{,}768}{48\text{,}939\text{,}635\text{,}836\text{,}927\text{,}856\text{,}157\text{,}843\text{,}580\text{,}875}:=C_{3}, \\& \lim_{n\rightarrow\infty}n^{23} \bigl(r_{4}(n)-\gamma \bigr) \\& \quad =\frac {10\text{,}081\text{,}089\text{,}508\text{,}192\text{,}698\text{,}621\text{,}180\text{,}096\text{,}244\text{,}326\text{,}317\text{,}115\text{,}783\text{,}678\text{,}789\text{,}502\text{,}403}{ 1\text{,}210\text{,}912\text{,}402\text{,}904\text{,}902\text{,}219\text{,}665\text{,}371\text{,}334\text{,}928\text{,}392\text{,}072\text{,}785\text{,}413\text{,}652\text{,}121\text{,}600} \\& \quad :=C_{4}, \\& \lim_{n\rightarrow\infty}n^{27} \bigl(r_{5}(n)-\gamma \bigr)=\frac {587\ldots212}{890\ldots625}:=C_{5}. \end{aligned}$$

(1.5)

Furthermore, for $r_{2}(n)$ and $r_{3}(n)$, we also have the following inequalities.

Theorem 2

Let $r_{2}(n)$, $r_{3}(n)$, $C_{2}$ and $C_{3}$ be defined in Theorem  1, then

$$\begin{aligned}& C_{2}\frac{1}{ (n+\frac{3}{2} )^{14}} < r_{2}(n)-\gamma< C_{2} \frac {1}{n^{14}}, \end{aligned}$$

(1.6)

$$\begin{aligned}& C_{3}\frac{1}{ (n+\frac{3}{2} )^{18}} < r_{3}(n)-\gamma< C_{3} \frac {1}{n^{18}}. \end{aligned}$$

(1.7)

Remark 1

In fact, Theorem 2 implies that $r_{2}(n)$ and $r_{3}(n)$ are strictly increasing functions of n. Certainly, it has similar inequalities for $r_{l}(n)$ ($4\le k\le5$), we omit these details. It should also be noted that (1.4) cannot deduce the monotony of $r_{3}(n)$.

Remark 2

It is worth pointing out that Theorem 2 provides sharp bounds and faster rate of convergence for harmonic sequence, which are superior to Theorems 3 and 4 in Mortici and Chen [5].

2 The proof of Theorem 1

The following lemma gives a method for measuring the rate of convergence. This lemma was first used by Mortici [14–18] for constructing asymptotic expansions, or to accelerate some convergences.

Lemma 1

If the sequence $(x_{n})_{n\in\mathbb{N}}$ is convergent to zero and there exists the limit

$$ \lim_{n\rightarrow+\infty}n^{s}(x_{n}-x_{n+1})=l \in[-\infty,+\infty] $$

(2.1)

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

$$ \lim_{n\rightarrow+\infty}n^{s-1}x_{n}= \frac{l}{s-1}. $$

(2.2)

In the sequel, we always assume $n\ge2$.

Based on our previous works [19–22], we will apply multiple-correction method to study faster convergence problem for constants of Euler-Mascheroni. In this paper, we always assume that the following conditions hold.

Condition 1

The initial-correction function $\eta_{0}(n)$ satisfies

$$\begin{aligned}& \lim_{n\rightarrow\infty} \bigl(v(n)-\eta_{0}(n) \bigr)=0, \\& \lim_{n\rightarrow\infty}n^{l_{0}} \bigl(v(n)-v(n+1)- \eta_{0}(n)+\eta_{0}(n+1) \bigr)=C_{0}\neq0, \end{aligned}$$

with some a positive integer $l\ge2$.

Condition 2

The kth correction function $\eta_{k}(n)$ has the form of $-\frac{C_{k-1}}{\Phi_{k}(l_{k-1};n)}$, where

$$ \lim_{n\rightarrow\infty}n^{l_{k-1}} \Biggl(v(n)-v(n+1)- \sum _{j=0}^{k-1} \bigl(\eta_{j}(n)- \eta_{j}(n+1) \bigr) \Biggr)=C_{k-1}\neq0. $$

Condition 3

The function $v(x)$ satisfies $v(x)\in C^{\infty}[1,+\infty)$.

Step 1

(The initial-correction)

We choose $\eta_{0}(n)=0$, and let

$$ r_{0}(n):=\sum_{m=1}^{n} \frac{1}{m} -\frac{1}{2}\ln \bigl(n^{2}+b n+c \bigr)- \eta_{0}(n). $$

(2.3)

Developing expression (2.3) into power series expansion in $\frac{1}{n}$, we obtain

$$ r_{0}(n)-r_{0}(n+1)=\frac{1-b}{2} \frac{1}{n^{2}}+\frac {3b^{2}+3b-6c-4}{6}\frac{1}{n^{3}}+O \biggl( \frac{1}{n^{4}} \biggr). $$

(2.4)

By Lemma 1, we have

• (i) If $b\neq1$ and $c\neq\frac{1}{3}$, then the rate of convergence of $(r_{0}(n)-\gamma )_{n\in\mathbb{N}}$ is $n^{-1}$ since

  $$ \lim_{n\rightarrow\infty}n \bigl(r_{0}(n)-\gamma \bigr)= \frac {1-b}{2}\neq0. $$

• (ii) If $b=1$ and $c=\frac{1}{3}$, from (2.4) we have

  $$ r_{0}(n)-r_{0}(n+1)=-\frac{1}{45} \frac{1}{n^{5}}+O \biggl(\frac{1}{n^{6}} \biggr). $$

Hence the rate of convergence of $(r_{1}(n)-\gamma )_{n\in\mathbb{N}}$ is $n^{-5}$ since

$$ \lim_{n\rightarrow\infty}n^{4} \bigl(r_{0}(n)-\gamma \bigr)=-\frac{1}{180}. $$

Step 2

(The first-correction)

We let

$$ \eta_{1}(n)=\frac{a_{1}}{(n+\frac{1}{2})^{4}+b_{2} (n+\frac{1}{2})^{2}+b_{0}} $$

(2.5)

and define

$$ r_{1}(n):=\sum_{m=1}^{n} \frac{1}{m} -\frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr)-\eta_{0}(n)-\eta_{1}(n). $$

(2.6)

By the same method as above, we find $a_{1}=-\frac{1}{180}$, $b_{2}=\frac {85}{126}$, $b_{0}=-\frac{18\text{,}287}{63\text{,}504}$.

Applying Lemma 1 again, one has

$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{11} \bigl(r_{1}(n)-r_{1}(n+1) \bigr)=\frac{915\text{,}056}{24\text{,}754\text{,}653}, \end{aligned}$$

(2.7)

$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{10} \bigl(r_{1}(n)-\gamma \bigr)=\frac{457\text{,}528}{123\text{,}773\text{,}265}. \end{aligned}$$

(2.8)

Step 3

(The second-correction)

Similarly, we set the second-correction function in the form of

$$ \eta_{2}(n)=\frac{a_{2}}{(n+\frac{1}{2})^{2}+k_{2}} $$

(2.9)

and define

$$\begin{aligned} r_{2}(n) :=&\sum_{m=1}^{n} \frac{1}{m} -\frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &{}-\frac{a_{1}}{(n+\frac{1}{2})^{4}+b_{2} (n+\frac{1}{2})^{2}+b_{0}+}\frac {a_{2}}{(n+\frac{1}{2})^{2}+k_{2}}. \end{aligned}$$

(2.10)

By the same method as above, we find $a_{2}=\frac{1\text{,}830\text{,}112}{2\text{,}750\text{,}517}$, $k_{2}=\frac{19\text{,}949\text{,}142\text{,}781}{5\text{,}995\text{,}446\text{,}912}$.

Applying Lemma 1 again, one has

$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{15} \bigl(r_{2}(n)-r_{2}(n+1) \bigr)=\frac{16\text{,}615\text{,}600\text{,}105\text{,}955}{79\text{,}366\text{,}013\text{,}236\text{,}224}, \end{aligned}$$

(2.11)

$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{14} \bigl(r_{2}(n)-\gamma \bigr)=\frac{16\text{,}615\text{,}600\text{,}105\text{,}955}{1\text{,}111\text{,}124\text{,}185\text{,}307\text{,}136}. \end{aligned}$$

(2.12)

Repeat the above approach to determine $a_{3}$ to $a_{5}$ step by step. However, the computations become very difficult to compute $a_{l}$ and $k_{l}$, $l>5$. In this paper we will use the Mathematica software to manipulate symbolic computations.

This completes the proof of Theorem 1.

3 The proof of Theorem 2

The following lemma plays an important role in the proof of our inequalities, which is a direct consequence of the Hermite-Hadamard inequality.

Lemma 2

Let $f''(x)$ be a continuous function. If $f''(x)>0$, then

$$ \int_{a}^{a+1}f(x) \, dx > f(a+1/2). $$

(3.1)

In the sequel, the notation $P_{k}(x)$ means a polynomial of degree k in x with all of its non-zero coefficients positive, which may be different at each occurrence.

Let us begin to prove Theorem 2. Note $r_{2}(\infty)= \gamma$, it is easy to see

$$ r_{2}(n)-\gamma=\sum_{m=n}^{\infty} \bigl(r_{2}(m)-r_{2}(m+1) \bigr) =\sum _{m=n}^{\infty}f(m), $$

(3.2)

where

$$\begin{aligned} f(m) =&-\frac{1}{m+1}+\frac{1}{2}\ln\frac{(m+1)^{2}+(m+1)+\frac{1}{3}}{m^{2}+m+\frac{1}{3}} \\ &{}+\frac{a_{1}}{(m+\frac{3}{2})^{4}+b_{2} (m+\frac{3}{2})^{2}+b_{0}+}\frac {a_{2}}{(m+\frac{3}{2})^{2}+k_{2}} \\ &{}-\frac{a_{1}}{(m+\frac{1}{2})^{4}+b_{2} (m+\frac{1}{2})^{2}+b_{0}+}\frac{a_{2}}{(m+\frac{1}{2})^{2}+k_{2}}. \end{aligned}$$

Let $D_{1}=\frac{83\text{,}078\text{,}000\text{,}529\text{,}775}{26\text{,}455\text{,}337\text{,}745\text{,}408}$. By using the Mathematica software, we have

$$\begin{aligned}& f'(x)+D_{1}\frac{1}{ (x+\frac{3}{2} )^{16}} \\& \quad =-\frac {P_{29}(x)}{904\text{,}235\text{,}176\text{,}845(1+x)^{2}(3+2x)^{16} (1+3x+3x^{2})(7+9x+3x^{2}){P_{6}^{(1)}}^{2}(x){P_{6}^{(2)}}^{2}(x)}>0 \end{aligned}$$

and

$$\begin{aligned}& f'(x)+D_{1}\frac{1}{ (x+\frac{1}{2} )^{16}} \\& \quad =\frac {P_{29}(x)}{904\text{,}235\text{,}176\text{,}845(1+x)^{2}(1+2x)^{16} (1+3x+3x^{2})(7+9x+3x^{2}){P_{8}^{(1)}}^{2}(x){P_{8}^{(2)}}^{2}(x)}>0. \end{aligned}$$

Hence, we get the following inequalities for $x\ge1$:

$$ D_{1}\frac{1}{(x+\frac{3}{2})^{16}}< -f'(x)< D_{1} \frac{1}{(x+\frac{1}{2})^{16}}. $$

(3.3)

Applying $f(\infty)=0$, (3.3) and Lemma 2, we get

$$\begin{aligned} f(m) =& -\int_{m}^{\infty}f'(x) \, dx \le D_{1}\int_{m}^{\infty} \biggl( x+ \frac{1}{2} \biggr)^{-16}\, dx \\ =&\frac{D_{1}}{15} \biggl(m+\frac{1}{2} \biggr)^{-15} \le \frac{D_{1}}{15}\int_{m}^{m+1}x^{-15}\, dx. \end{aligned}$$

(3.4)

From (3.1) and (3.4) we obtain

$$\begin{aligned} r_{2}(n)-\gamma \le& \sum_{m=n}^{\infty} \frac{D_{1}}{15}\int_{m}^{m+1}x^{-15}\, dx \\ =&\frac{D_{1}}{15}\int_{n}^{\infty}x^{-15} \, dx =\frac{D_{1}}{210}\frac{1}{n^{14}}=C_{2}\frac{1}{n^{14}}. \end{aligned}$$

(3.5)

Similarly, we also have

$$\begin{aligned} f(m) =& -\int_{m}^{\infty}f'(x)\, dx \ge D_{1}\int_{m}^{\infty} \biggl( x+ \frac{3}{2} \biggr)^{-16}\, dx \\ =&\frac{D_{1}}{15} \biggl(m+\frac{3}{2} \biggr)^{-15} \ge \frac{D_{1}}{19}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-15}\, dx \end{aligned}$$

and

$$\begin{aligned} r_{2}(n)-\gamma \ge& \sum_{m=n}^{\infty} \frac{D_{1}}{15}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-15} \, dx \\ =&\frac{D_{1}}{15}\int_{n+\frac{3}{2}}^{\infty}x^{-15} \, dx =\frac{D_{1}}{210}\frac{1}{ (n+\frac{3}{2} )^{14}}=C_{2}\frac {1}{ (n+\frac{3}{2} )^{14}}. \end{aligned}$$

(3.6)

Combining (3.5) and (3.6) completes the proof of (1.6).

Note $r_{3}(\infty)=\gamma$, it is easy to see

$$ r_{3}(n)-\gamma=\sum_{m=n}^{\infty} \bigl(r_{3}(m)-r_{3}(m+1) \bigr) =\sum _{m=n}^{\infty}g(m), $$

(3.7)

where

$$\begin{aligned} g(m) =&\frac{1}{m+1}-\frac{1}{2}\ln\frac{(m+1)^{2}+(m+1)+\frac{1}{3}}{m^{2}+m+\frac{1}{3}} \\ &{}-\frac{a_{1}}{(m+\frac{3}{2})^{4}+b_{2} (m+\frac{3}{2})^{2}+b_{0}+}\frac {a_{2}}{(m+\frac{3}{2})^{2}+k_{2}+}\frac{a_{3}}{(m+\frac{3}{2})^{2}+k_{3}} \\ &{}+\frac{a_{1}}{(m+\frac{1}{2})^{4}+b_{2} (m+\frac{1}{2})^{2}+b_{0}+}\frac {a_{2}}{(m+\frac{1}{2})^{2}+k_{2}+}\frac{a_{3}}{(m+\frac{1}{2})^{2}+k_{3}}. \end{aligned}$$

Let $D_{2}=\frac {21\text{,}655\text{,}539\text{,}661\text{,}060\text{,}973\text{,}661\text{,}932\text{,}007\text{,}536}{286\text{,}196\text{,}700\text{,}800\text{,}747\text{,}696\text{,}829\text{,}494\text{,}625}$. By using the Mathematica software, we have

$$\begin{aligned}& g'(x)+D_{2}\frac{1}{ (x+\frac{3}{2} )^{20}} \\& \quad =-\frac {P_{37}(x)}{286\ldots625(1+x)^{2}(3+2x)^{20} (1+3x+3x^{2})(7+9x+3x^{2}){P_{8}^{(1)}}^{2}(x){P_{8}^{(2)}}^{2}(x)}< 0 \end{aligned}$$

and

$$\begin{aligned}& g'(x)+D_{2}\frac{1}{ (x+\frac{1}{2} )^{20}} \\& \quad =\frac {P_{37}(x)}{286\ldots625(1+x)^{2}(1+2x)^{20} (1+3x+3x^{2})(7+9x+3x^{2}){P_{8}^{(1)}}^{2}(x){P_{8}^{(2)}}^{2}(x)}>0. \end{aligned}$$

Hence, we get the following inequalities for $x\ge1$:

$$ D_{2}\frac{1}{(x+\frac{3}{2})^{20}}< -g'(x)< D_{2} \frac{1}{(x+\frac{1}{2})^{20}}. $$

(3.8)

Applying $g(\infty)=0$, (3.8) and Lemma 2, we get

$$\begin{aligned} g(m) =& -\int_{m}^{\infty}g'(x) \, dx \le D_{2}\int_{m}^{\infty} \biggl( x+ \frac{1}{2} \biggr)^{-20} \, dx \\ =&\frac{D_{2}}{19} \biggl(m+\frac{1}{2} \biggr)^{-19} \le \frac{D_{2}}{19}\int_{m}^{m+1}x^{-19} \, dx. \end{aligned}$$

(3.9)

From (3.1) and (3.4) we obtain

$$\begin{aligned} r_{3}(n)-\gamma \le& \sum_{m=n}^{\infty} \frac{D_{2}}{19}\int_{m}^{m+1}x^{-19} \, dx \\ =&\frac{D_{2}}{19}\int_{n}^{\infty}x^{-19} \, dx =\frac{D_{2}}{342}\frac{1}{n^{18}}=C_{3}\frac{1}{n^{18}}. \end{aligned}$$

(3.10)

Similarly, we also have

$$\begin{aligned} g(m) =& -\int_{m}^{\infty}g'(x)\, dx \ge D_{2}\int_{m}^{\infty} \biggl( x+ \frac{3}{2} \biggr)^{-20} \, dx \\ =&\frac{D_{2}}{19} \biggl(m+\frac{3}{2} \biggr)^{-19} \ge \frac{D_{2}}{19}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-19}\, dx \end{aligned}$$

and

$$\begin{aligned} r_{3}(n)-\gamma \ge& \sum_{m=n}^{\infty} \frac{D_{2}}{19}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-19} \, dx \\ =&\frac{D_{2}}{19}\int_{n+\frac{3}{2}}^{\infty}x^{-19} \, dx =\frac{D_{2}}{342}\frac{1}{ (n+\frac{3}{2} )^{18}}=C_{3}\frac {1}{ (n+\frac{3}{2} )^{18}}. \end{aligned}$$

(3.11)

Combining (3.5) and (3.6) completes the proof of (1.7).