Open Access

Some sharp continued fraction inequalities for the Euler-Mascheroni constant

Journal of Inequalities and Applications20152015:308

Received: 28 July 2015

Accepted: 23 September 2015

Published: 30 September 2015


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


Euler-Mascheroni constant rate of convergence continued fraction multiple-correction


11Y60 41A25 41A20

1 Introduction

Euler introduced a constant, then later this constant was called ‘Euler’s constant’ as the limit of the sequence
$$ \gamma(n):=\sum_{m=1}^{n} \frac{1}{m} -\ln n. $$
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, 413] 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
$$ \lim_{n\rightarrow\infty}n^{12} \bigl(\nu(n)-\gamma \bigr)=-\frac {796\text{,}801}{43\text{,}783\text{,}740}. $$
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
$$\begin{aligned}& \frac{1}{72(n+1)^{3}}< \gamma-r_{2}(n)< \frac{1}{72n^{3}}, \end{aligned}$$
$$\begin{aligned}& \frac{1}{120(n+1)^{4}}< r_{3}(n)-\gamma< \frac{1}{120(n-1)^{4}}. \end{aligned}$$

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:
$$\begin{aligned} r_{k}(n) =&\sum_{m=1}^{n} \frac{1}{m}-\frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &{}-\frac{a_{1}}{(n+\frac{1}{2})^{4}+b_{2} (n+\frac{1}{2})^{2}+b_{0}+}\frac {a_{2}}{(n+\frac{1}{2})^{2}+k_{2}+}\ddots, \end{aligned}$$
$$\begin{aligned}& b_{2}=\frac{85}{126}, \qquad b_{0}=-\frac{18\text{,}287}{63\text{,}504}; \\& a_{1}=-\frac{1}{180}, \qquad a_{2}=\frac{1\text{,}830\text{,}112}{2\text{,}750\text{,}517}, \qquad a_{3}=-\frac {36\text{,}637\text{,}398\text{,}233\text{,}630\text{,}775}{9\text{,}056\text{,}534\text{,}057\text{,}598\text{,}976}, \\& a_{4}=-\frac {16\text{,}486\text{,}938\text{,}208\text{,}076\text{,}386\text{,}182\text{,}240\text{,}455\text{,}433\text{,}101\text{,}197\text{,}312}{1\text{,}114\text{,}333\text{,}428\text{,}433\text{,}648\text{,}110\text{,}295\text{,}680\text{,}727\text{,}490\text{,}570\text{,}625}, \\& a_{5}=-\frac{202\ldots665}{538\ldots376};\qquad \ldots, \\& k_{2}=\frac{19\text{,}949\text{,}142\text{,}781}{5\text{,}995\text{,}446\text{,}912},\qquad k_{3}=\frac {83\text{,}116\text{,}192\text{,}006\text{,}963\text{,}596\text{,}639\text{,}745\text{,}097}{13\text{,}306\text{,}111\text{,}671\text{,}966\text{,}702\text{,}431\text{,}356\text{,}800}, \\& k_{4}=\frac {3\text{,}223\text{,}193\text{,}895\text{,}482\text{,}188\text{,}285\text{,}536\text{,}076\text{,}617\text{,}166\text{,}535\text{,}131\text{,}877\text{,}815\text{,}854\text{,}443}{314\text{,}899\text{,}884\text{,}866\text{,}916\text{,}635\text{,}406\text{,}301\text{,}198\text{,}738\text{,}123\text{,}952\text{,}960\text{,}626\text{,}300\text{,}800}, \\& k_{5}=\frac{155\ldots 06831}{102\ldots440},\qquad \ldots. \end{aligned}$$
For \(1\le l\le5\), we have
$$\begin{aligned}& \lim_{n\rightarrow\infty}n^{11} \bigl(r_{1}(n)- \gamma \bigr)=\frac {457\text{,}528}{123\text{,}773\text{,}265}:=C_{1}, \\& \lim_{n\rightarrow\infty}n^{15} \bigl(r_{2}(n)-\gamma \bigr)=\frac {16\text{,}615\text{,}600\text{,}105\text{,}955}{1\text{,}111\text{,}124\text{,}185\text{,}307\text{,}136}:=C_{2}, \\& \lim_{n\rightarrow\infty}n^{19} \bigl(r_{3}(n)-\gamma \bigr)=\frac {10\text{,}827\text{,}769\text{,}830\text{,}530\text{,}486\text{,}830\text{,}966\text{,}003\text{,}768}{48\text{,}939\text{,}635\text{,}836\text{,}927\text{,}856\text{,}157\text{,}843\text{,}580\text{,}875}:=C_{3}, \\& \lim_{n\rightarrow\infty}n^{23} \bigl(r_{4}(n)-\gamma \bigr) \\& \quad =\frac {10\text{,}081\text{,}089\text{,}508\text{,}192\text{,}698\text{,}621\text{,}180\text{,}096\text{,}244\text{,}326\text{,}317\text{,}115\text{,}783\text{,}678\text{,}789\text{,}502\text{,}403}{ 1\text{,}210\text{,}912\text{,}402\text{,}904\text{,}902\text{,}219\text{,}665\text{,}371\text{,}334\text{,}928\text{,}392\text{,}072\text{,}785\text{,}413\text{,}652\text{,}121\text{,}600} \\& \quad :=C_{4}, \\& \lim_{n\rightarrow\infty}n^{27} \bigl(r_{5}(n)-\gamma \bigr)=\frac {587\ldots212}{890\ldots625}:=C_{5}. \end{aligned}$$

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
$$\begin{aligned}& C_{2}\frac{1}{ (n+\frac{3}{2} )^{14}} < r_{2}(n)-\gamma< C_{2} \frac {1}{n^{14}}, \end{aligned}$$
$$\begin{aligned}& C_{3}\frac{1}{ (n+\frac{3}{2} )^{18}} < r_{3}(n)-\gamma< C_{3} \frac {1}{n^{18}}. \end{aligned}$$

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 [1418] 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
$$ \lim_{n\rightarrow+\infty}n^{s}(x_{n}-x_{n+1})=l \in[-\infty,+\infty] $$
with \(s>1\), then there exists the limit
$$ \lim_{n\rightarrow+\infty}n^{s-1}x_{n}= \frac{l}{s-1}. $$

In the sequel, we always assume \(n\ge2\).

Based on our previous works [1922], 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
$$ r_{0}(n):=\sum_{m=1}^{n} \frac{1}{m} -\frac{1}{2}\ln \bigl(n^{2}+b n+c \bigr)- \eta_{0}(n). $$
Developing expression (2.3) into power series expansion in \(\frac{1}{n}\), we obtain
$$ r_{0}(n)-r_{0}(n+1)=\frac{1-b}{2} \frac{1}{n^{2}}+\frac {3b^{2}+3b-6c-4}{6}\frac{1}{n^{3}}+O \biggl( \frac{1}{n^{4}} \biggr). $$
By Lemma 1, we have
  1. (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
    $$ \lim_{n\rightarrow\infty}n \bigl(r_{0}(n)-\gamma \bigr)= \frac {1-b}{2}\neq0. $$
  2. (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
$$ \eta_{1}(n)=\frac{a_{1}}{(n+\frac{1}{2})^{4}+b_{2} (n+\frac{1}{2})^{2}+b_{0}} $$
and define
$$ r_{1}(n):=\sum_{m=1}^{n} \frac{1}{m} -\frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr)-\eta_{0}(n)-\eta_{1}(n). $$
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
$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{11} \bigl(r_{1}(n)-r_{1}(n+1) \bigr)=\frac{915\text{,}056}{24\text{,}754\text{,}653}, \end{aligned}$$
$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{10} \bigl(r_{1}(n)-\gamma \bigr)=\frac{457\text{,}528}{123\text{,}773\text{,}265}. \end{aligned}$$

Step 3

(The second-correction)

Similarly, we set the second-correction function in the form of
$$ \eta_{2}(n)=\frac{a_{2}}{(n+\frac{1}{2})^{2}+k_{2}} $$
and define
$$\begin{aligned} r_{2}(n) :=&\sum_{m=1}^{n} \frac{1}{m} -\frac{1}{2}\ln \biggl(n^{2}+n+\frac{1}{3} \biggr) \\ &{}-\frac{a_{1}}{(n+\frac{1}{2})^{4}+b_{2} (n+\frac{1}{2})^{2}+b_{0}+}\frac {a_{2}}{(n+\frac{1}{2})^{2}+k_{2}}. \end{aligned}$$
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
$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{15} \bigl(r_{2}(n)-r_{2}(n+1) \bigr)=\frac{16\text{,}615\text{,}600\text{,}105\text{,}955}{79\text{,}366\text{,}013\text{,}236\text{,}224}, \end{aligned}$$
$$\begin{aligned}& \lim_{n\rightarrow\infty} n^{14} \bigl(r_{2}(n)-\gamma \bigr)=\frac{16\text{,}615\text{,}600\text{,}105\text{,}955}{1\text{,}111\text{,}124\text{,}185\text{,}307\text{,}136}. \end{aligned}$$

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

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
$$ r_{2}(n)-\gamma=\sum_{m=n}^{\infty} \bigl(r_{2}(m)-r_{2}(m+1) \bigr) =\sum _{m=n}^{\infty}f(m), $$
$$\begin{aligned} f(m) =&-\frac{1}{m+1}+\frac{1}{2}\ln\frac{(m+1)^{2}+(m+1)+\frac{1}{3}}{m^{2}+m+\frac{1}{3}} \\ &{}+\frac{a_{1}}{(m+\frac{3}{2})^{4}+b_{2} (m+\frac{3}{2})^{2}+b_{0}+}\frac {a_{2}}{(m+\frac{3}{2})^{2}+k_{2}} \\ &{}-\frac{a_{1}}{(m+\frac{1}{2})^{4}+b_{2} (m+\frac{1}{2})^{2}+b_{0}+}\frac{a_{2}}{(m+\frac{1}{2})^{2}+k_{2}}. \end{aligned}$$
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
$$\begin{aligned}& f'(x)+D_{1}\frac{1}{ (x+\frac{3}{2} )^{16}} \\& \quad =-\frac {P_{29}(x)}{904\text{,}235\text{,}176\text{,}845(1+x)^{2}(3+2x)^{16} (1+3x+3x^{2})(7+9x+3x^{2}){P_{6}^{(1)}}^{2}(x){P_{6}^{(2)}}^{2}(x)}>0 \end{aligned}$$
$$\begin{aligned}& f'(x)+D_{1}\frac{1}{ (x+\frac{1}{2} )^{16}} \\& \quad =\frac {P_{29}(x)}{904\text{,}235\text{,}176\text{,}845(1+x)^{2}(1+2x)^{16} (1+3x+3x^{2})(7+9x+3x^{2}){P_{8}^{(1)}}^{2}(x){P_{8}^{(2)}}^{2}(x)}>0. \end{aligned}$$
Hence, we get the following inequalities for \(x\ge1\):
$$ D_{1}\frac{1}{(x+\frac{3}{2})^{16}}< -f'(x)< D_{1} \frac{1}{(x+\frac{1}{2})^{16}}. $$
Applying \(f(\infty)=0\), (3.3) and Lemma 2, we get
$$\begin{aligned} f(m) =& -\int_{m}^{\infty}f'(x) \, dx \le D_{1}\int_{m}^{\infty} \biggl( x+ \frac{1}{2} \biggr)^{-16}\, dx \\ =&\frac{D_{1}}{15} \biggl(m+\frac{1}{2} \biggr)^{-15} \le \frac{D_{1}}{15}\int_{m}^{m+1}x^{-15}\, dx. \end{aligned}$$
From (3.1) and (3.4) we obtain
$$\begin{aligned} r_{2}(n)-\gamma \le& \sum_{m=n}^{\infty} \frac{D_{1}}{15}\int_{m}^{m+1}x^{-15}\, dx \\ =&\frac{D_{1}}{15}\int_{n}^{\infty}x^{-15} \, dx =\frac{D_{1}}{210}\frac{1}{n^{14}}=C_{2}\frac{1}{n^{14}}. \end{aligned}$$
Similarly, we also have
$$\begin{aligned} f(m) =& -\int_{m}^{\infty}f'(x)\, dx \ge D_{1}\int_{m}^{\infty} \biggl( x+ \frac{3}{2} \biggr)^{-16}\, dx \\ =&\frac{D_{1}}{15} \biggl(m+\frac{3}{2} \biggr)^{-15} \ge \frac{D_{1}}{19}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-15}\, dx \end{aligned}$$
$$\begin{aligned} r_{2}(n)-\gamma \ge& \sum_{m=n}^{\infty} \frac{D_{1}}{15}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-15} \, dx \\ =&\frac{D_{1}}{15}\int_{n+\frac{3}{2}}^{\infty}x^{-15} \, dx =\frac{D_{1}}{210}\frac{1}{ (n+\frac{3}{2} )^{14}}=C_{2}\frac {1}{ (n+\frac{3}{2} )^{14}}. \end{aligned}$$
Combining (3.5) and (3.6) completes the proof of (1.6).
Note \(r_{3}(\infty)=\gamma\), it is easy to see
$$ r_{3}(n)-\gamma=\sum_{m=n}^{\infty} \bigl(r_{3}(m)-r_{3}(m+1) \bigr) =\sum _{m=n}^{\infty}g(m), $$
$$\begin{aligned} g(m) =&\frac{1}{m+1}-\frac{1}{2}\ln\frac{(m+1)^{2}+(m+1)+\frac{1}{3}}{m^{2}+m+\frac{1}{3}} \\ &{}-\frac{a_{1}}{(m+\frac{3}{2})^{4}+b_{2} (m+\frac{3}{2})^{2}+b_{0}+}\frac {a_{2}}{(m+\frac{3}{2})^{2}+k_{2}+}\frac{a_{3}}{(m+\frac{3}{2})^{2}+k_{3}} \\ &{}+\frac{a_{1}}{(m+\frac{1}{2})^{4}+b_{2} (m+\frac{1}{2})^{2}+b_{0}+}\frac {a_{2}}{(m+\frac{1}{2})^{2}+k_{2}+}\frac{a_{3}}{(m+\frac{1}{2})^{2}+k_{3}}. \end{aligned}$$
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
$$\begin{aligned}& g'(x)+D_{2}\frac{1}{ (x+\frac{3}{2} )^{20}} \\& \quad =-\frac {P_{37}(x)}{286\ldots625(1+x)^{2}(3+2x)^{20} (1+3x+3x^{2})(7+9x+3x^{2}){P_{8}^{(1)}}^{2}(x){P_{8}^{(2)}}^{2}(x)}< 0 \end{aligned}$$
$$\begin{aligned}& g'(x)+D_{2}\frac{1}{ (x+\frac{1}{2} )^{20}} \\& \quad =\frac {P_{37}(x)}{286\ldots625(1+x)^{2}(1+2x)^{20} (1+3x+3x^{2})(7+9x+3x^{2}){P_{8}^{(1)}}^{2}(x){P_{8}^{(2)}}^{2}(x)}>0. \end{aligned}$$
Hence, we get the following inequalities for \(x\ge1\):
$$ D_{2}\frac{1}{(x+\frac{3}{2})^{20}}< -g'(x)< D_{2} \frac{1}{(x+\frac{1}{2})^{20}}. $$
Applying \(g(\infty)=0\), (3.8) and Lemma 2, we get
$$\begin{aligned} g(m) =& -\int_{m}^{\infty}g'(x) \, dx \le D_{2}\int_{m}^{\infty} \biggl( x+ \frac{1}{2} \biggr)^{-20} \, dx \\ =&\frac{D_{2}}{19} \biggl(m+\frac{1}{2} \biggr)^{-19} \le \frac{D_{2}}{19}\int_{m}^{m+1}x^{-19} \, dx. \end{aligned}$$
From (3.1) and (3.4) we obtain
$$\begin{aligned} r_{3}(n)-\gamma \le& \sum_{m=n}^{\infty} \frac{D_{2}}{19}\int_{m}^{m+1}x^{-19} \, dx \\ =&\frac{D_{2}}{19}\int_{n}^{\infty}x^{-19} \, dx =\frac{D_{2}}{342}\frac{1}{n^{18}}=C_{3}\frac{1}{n^{18}}. \end{aligned}$$
Similarly, we also have
$$\begin{aligned} g(m) =& -\int_{m}^{\infty}g'(x)\, dx \ge D_{2}\int_{m}^{\infty} \biggl( x+ \frac{3}{2} \biggr)^{-20} \, dx \\ =&\frac{D_{2}}{19} \biggl(m+\frac{3}{2} \biggr)^{-19} \ge \frac{D_{2}}{19}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-19}\, dx \end{aligned}$$
$$\begin{aligned} r_{3}(n)-\gamma \ge& \sum_{m=n}^{\infty} \frac{D_{2}}{19}\int_{m+\frac{3}{2}}^{m+\frac{5}{2}}x^{-19} \, dx \\ =&\frac{D_{2}}{19}\int_{n+\frac{3}{2}}^{\infty}x^{-19} \, dx =\frac{D_{2}}{342}\frac{1}{ (n+\frac{3}{2} )^{18}}=C_{3}\frac {1}{ (n+\frac{3}{2} )^{18}}. \end{aligned}$$
Combining (3.5) and (3.6) completes the proof of (1.7).



The authors thank the referees for their careful reading of the manuscript and insightful comments. Research of this paper was supported by the National Natural Science Foundation of China (Grant Nos. 61403034, 11171014 and 11571267), and Beijing Municipal Commission of Education Science and Technology Program KM201310017006.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

School of Mathematics and System Science, Beihang University
Department of Mathematics and Physics, Beijing Institute of Petrochemical Technology


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