Dedekind和是Dedekind在研究$ \eta $函数时引入的一种和式: 设$ k>0 $, $ h $为任意整数, 经典的Dedekind和[1] $ S\left ( h, k \right ) $被定义为:

$ \begin{equation*} S\left ( h,k \right ) =\sum\limits_{a=1}^{k}\left ( \left ( \frac{a}{k} \right ) \right ) \left ( \left ( \frac{ah}{k} \right ) \right ) , \end{equation*} $

$ \left ( \left ( x \right ) \right ) $是分段函数:

$ \begin{equation*} {\left ( \left ( x \right ) \right )=} \begin{cases} x-\left [ x \right ] -\frac{1}{2}, \quad & x \notin Z, \\ 0, \quad & x \in Z, \end{cases} \end{equation*} $

$ \left [ x\right ] $是不超过$ x $的最大整数, $ Z $是整数集. 许多学者对其进行了不同角度的研究, 见[2-9]. 有关于Dedekind和最著名也是最重要的性质之一是互反公式, 即当$ h, k > 0 $且$ \left ( h,k \right )=1 $时有:

$ \begin{equation*} S\left ( h, k \right ) +S\left ( k, h \right ) =\frac{h^{2} +k^{2} +1}{12hk}-\frac{1}{4}. \end{equation*} $

当$ n\ge 2 $时, 一般线性递推数列Fibonacci$ \left \{ F_{n} \right \} $数列和Lucas $ \left \{ L_{n} \right \} $数列, 由以下关系式给出:

$ \begin{equation*} F_{0} =0, \quad F_{1} =1, \quad F_{n} = F_{n-1} +F_{n-2}, \quad n\ge 2, \end{equation*} $

$ \begin{equation*} L_{0} =2, \quad L_{1} =1, \quad L_{n} = L_{n-1} +L_{n-2}, \quad n\ge 2. \end{equation*} $

在[10]中, 学者推广线性递推数列为非线性递推数列, 即双周期Fibonacci $ \left \{ f_{n} \right \} $数列, 其定义如下:

$ \begin{equation} f_{0}=0, \quad f_{1}=1, \quad {f_{n} =} \begin{cases} af_{n-1} +f_{n-2}, \quad & n \equiv 0 \pmod{2},\\ bf_{n-1} +f_{n-2}, \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2 , \end{equation} $

(1.1)

其中$ a $, $ b $为任意非负实数且$ a\ge1 $, $ b\ge1 $. 显然当$ a=b=1 $时, 双周期Fibonacci$ \left \{ f_{n} \right \} $数列变为Fibonacci $ \left \{F_{n} \right \} $数列. 当$ a=b=k $时, 双周期Fibonacci $ \left \{ f_{n} \right \} $数列变为$ k $-Fibonacci $ \left \{ Q_{n} \right \} $数列[11].

同样地, 在[12] 中, 学者介绍了双周期Lucas $ \left \{ l_{n} \right \} $数列, 定义如下:

$ \begin{equation} l_{0} =2, \quad l_{1} =a , \quad {l_{n} =} \begin{cases} bl_{n-1} +l_{n-2} , \quad & n \equiv 0 \pmod{2},\\ a l_{n-1} +l_{n-2} , \quad &n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation} $

(1.2)

其中$ a $, $ b $为任意非负实数且$ a\ge1 $, $ b\ge1 $. 且当$ a=b=1 $时, 双周期Lucas$ \left \{ l_{n} \right \} $数列变为Lucas $ \left \{L_{n} \right \} $数列. 当$ a=b=k $时, 双周期Lucas $ \left \{ l_{n} \right \} $数列变为$ k $-Lucas $ \left \{ P_{n} \right \} $数列[13].

我们给出$ \left \{f_{n} \right \} $和$ \left \{l_{n} \right \} $的负下标表达式如下:

$ \begin{equation} f_{-n} =\left ( -1 \right )^{n+1}f_{n}, \quad l_{-n} =\left ( -1 \right )^{n}l_{n}. \end{equation} $

(1.3)

另外定义分段函数$ \zeta \left ( n \right ) $如下:

$ \begin{equation} {\zeta \left ( n \right )=} \begin{cases} 0, \quad & n \equiv 0 \pmod{2}, \\ 1, \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2. \end{equation} $

(1.4)

在[10] 和[12]中, 学者们分别得到$ \left \{ f_{n} \right \} $和$ \left \{ l_{n} \right \} $的Binet公式如下:

$ \begin{equation} f_{n} =\frac{a^{\zeta \left (n+1 \right) }}{ \left(ab \right )^{\left [\frac{n}{2} \right ] } }\left (\frac{\alpha^{n} -\beta ^{n} }{\alpha - \beta }\right), \end{equation} $

(1.5)

$ \begin{equation} l_{n} =\frac{a^{\zeta \left (n \right) }}{ \left(ab \right )^{\left [\frac{n+1}{2} \right ] } }\left (\alpha^{n} + \beta ^{n} \right), \end{equation} $

(1.6)

其中$ \alpha $和$ \beta $是方程$ \omega ^{2}-ab \omega -ab=0 $的根, 即$ \alpha=\frac{ab +\sqrt{a^{2} b^{2} +4ab } }{2} $且$ \beta =\frac{ab -\sqrt{a^{2}b^{2} +4ab } }{2} $. 我们得到以下关于$ \alpha $和$ \beta $的算术性质:

$ \begin{equation*} \alpha + \beta =ab, \quad \alpha - \beta =\sqrt{a^{2} b^{2} +4ab}, \quad \alpha \beta =-ab. \end{equation*} $

显然, 当$ a\ge 1 $及$ b\ge 1 $时, $ 1< \alpha $且$ -1< \beta <0 $. 更多有关双周期Fibonacci数列和双周期Lucas数列的讨论, 见[14-20].

另外根据Dedekind和的互反公式以及$ \left \{ f_{n} \right \} $与$ \left \{ l_{n} \right \} $的递推关系, 我们可以得到如下等式:

$ \begin{equation*} S\left ( f_{0} , f_{1} \right ) =S\left ( 0, 1 \right ) =0, \quad S\left ( f_{1} , f_{2} \right ) =S\left ( 1, a \right ) =\frac{\left ( a-1 \right )\left ( a-2 \right ) }{12a}. \end{equation*} $

当$ a $是奇数时:

$ \begin{equation*} S\left ( l_{0} , l_{1} \right ) =S\left ( 2, a \right ) = \frac{\left ( a-1 \right )\left ( a-5 \right ) }{24a}, \end{equation*} $

$ \begin{equation*} S\left ( l_{1} , l_{2} \right ) =S\left ( a, ab+2 \right ) = \frac{b\left ( 2ab+3-a^{2} \right ) }{24\left ( ab+2 \right ) } . \end{equation*} $

近年来, 许多学者都研究了涉及二阶线性递推数列的Dedekind和, 比如: 1999年, 张文鹏和易媛[21] 讨论了估计式$ \sum\limits_{n=1}^{m}S\left ( F_{n} , F_{n+1} \right ) $, 即当$ m $是正整数时有:

$ \begin{equation*} \sum\limits_{n=1}^{m}S\left ( F_{n} , F_{n+1} \right )=-\frac{\left ( \sqrt{5}-1 \right ) ^{2} }{48} m+C\left ( m \right )+ \mathcal{O}\left ( \frac{1}{\alpha ^{2m} } \right ), \end{equation*} $

其中, $ C\left ( m \right ) $是仅仅依赖于$ m $的常数, “$ \mathcal{O} $”表示Landau符号（对任意$ x\ge a $, $ g\left ( x \right ) > 0 $, 若$ f\left ( x \right ) /g\left ( x \right ) $有界, 则我们记作$ f\left ( x \right ) =\mathcal{O} \left ( g\left ( x \right ) \right ) $.

另外, 有学者[22] 得到了涉及$ k $-Fibonacci $ \left \{ Q_{n} \right \} $数列和$ k $-Lucas $ \left \{P_{n} \right \} $数列的Dedekind和的估计式. 更多涉及线性递推数列Dedekind和的研究见[23, 24].

本文研究了一类二阶非线性递推数列的Dedekind和, 即讨论了涉及双周期Fibonacci $ \left \{f_{n} \right \} $数列和双周期Lucas $ \left \{ l_{n} \right \} $数列的Dedekind和的估计式. 本文是对前人研究成果的推广, 对Dedekind和与递推数列的研究有很大的帮助. 本文的主要结果如下:

定理 1.1   记$ \left \{ f_{n} \right \} $是双周期Fibonacci数列, $ \left \{ l_{n} \right \} $是双周期Lucas数列, 令$ m $是正整数, 则有:

$ \begin{equation} \begin{split} & \sum\limits_{n=1}^{m} \left [ S\left ( f_{2n} ,f_{2n+1} \right )+ S\left ( f_{2n+1} ,f_{2n+2} \right ) \right ] =\frac{\left ( a\alpha -3\alpha +a+b \right )m }{12\alpha } \\ &\quad \quad \quad +\frac{1}{12} \sum\limits _{n=1}^{\infty }\left ( \frac{\beta ^{2n+2} }{\left ( ab \right )^{n+1} f_{2n+2} } + \frac{a\beta ^{2n+1} }{\left ( ab \right )^{n+1}f_{2n+1} } +\frac{ab+4}{l_{4n+3} +a} \right ) +\mathcal{O} \left ( \frac{\alpha ^{2m} }{\beta ^{2m} } \right ), \end{split} \end{equation} $

(1.7)

且当$ a $是奇数时,

$ \begin{equation} \begin{split} & \sum\limits_{n=1}^{m} \left [ S\left ( l_{2n} ,l_{2n+1} \right )+ S\left ( l_{2n+1} ,l_{2n+2} \right ) \right ] = \frac{\left ( b\alpha -3\alpha +a+b \right )m }{12\alpha } \\ &\quad \quad \quad+\frac{1}{12} \sum\limits _{n=1}^{\infty }\left ( \frac{\left ( \alpha -\beta \right ) }{\alpha }\left ( \frac{a\beta ^{2n+1} }{\left ( ab \right )^{n+1} l_{2n+2} } + \frac{\beta ^{2n} }{\left ( ab \right )^{n}l_{2n+1} } \right )+\frac{1}{l_{4n+3} -a} \right ) +\mathcal{O} \left ( \frac{\alpha ^{2m} }{\beta ^{2m} } \right ). \end{split} \end{equation} $

(1.8)

本节我们介绍几个引理, 以便定理的证明.

引理 2.1   记$ \left \{ f_{n} \right \} $是双周期Fibonacci数列, $ \left \{ l_{n} \right \} $是双周期Lucas数列, 则

$ \begin{equation} T\left ( f_{2n}, f_{2n+1}, f_{2n+2} \right )=\left ( \frac{a-3}{6} \right )\left ( \zeta \left ( n \right ) +\frac{\left ( -1 \right )^{n} }{2} \right ), \end{equation} $

(2.1)

且当$ a $是奇数时:

$ \begin{equation} T\left ( l_{2n}, l_{2n+1}, l_{2n+2} \right )=\left ( \frac{b-3}{6} \right )\left ( \zeta \left ( n \right ) +\frac{\left ( -1 \right )^{n} }{2} \right ), \end{equation} $

(2.2)

其中$ T\left ( x, y, z \right )=S\left ( x, y \right )+ S\left ( y, z \right )-\frac{x}{12y}-\frac{y}{12z}-\frac{1}{12yz} $.

证   在这里我们仅证明(2.1), (2.2)的证明类似. 根据$ \left \{ f_{n} \right \} $的递推关系与Dedekind的性质, 有

$ \begin{equation} \begin{split} &S\left ( f_{2n-1} ,f _{2n} \right ) +S\left ( f_{2n} ,f _{2n-1} \right ) +S\left ( f_{2n+1} ,f _{2n+2} \right ) +S\left ( f_{2n+2} ,f _{2n+1} \right )\\ =& S\left ( f_{2n-1} ,f _{2n} \right ) +S\left (af_{2n-1} +f_{2n-2} ,f _{2n-1} \right ) +S\left ( f_{2n+1} ,f _{2n+2} \right ) +S\left ( af_{2n+1} +f_{2n} ,f _{2n+1} \right ) \\ =& S\left ( f_{2n-1} ,f _{2n} \right ) +S\left ( f_{2n+1} ,f _{2n+2} \right ) +S\left ( f_{2n} ,f _{2n+1} \right ) +S\left ( f_{2n-2} ,f _{2n-1} \right ) . \end{split} \end{equation} $

(2.3)

又因为$ \left ( f_{2n } ,f _{2n+1} \right )=1 $, $ \left ( f_{2n+1} ,f _{2n+2} \right )=1 $. 由Dedekind和的互反公式, 我们很容易得到:

$ \begin{equation} \begin{split} &S\left ( f_{2n-1} ,f _{2n} \right ) +S\left ( f_{2n} ,f _{2n-1} \right ) +S\left ( f_{2n+1} ,f _{2n+2} \right ) +S\left ( f_{2n+2} ,f _{2n+1} \right )\\ =& \frac{1}{12f_{2n-1} f_{2n} }+\frac{1}{12f_{2n+1} f_{2n+2} } +\frac{f_{2n-1}}{12 f_{2n} }+\frac{f_{2n}}{12 f_{2n-1} } +\frac{f_{2n+1}}{12 f_{2n+2} }+\frac{f_{2n+2}}{12 f_{2n+1} }-\frac{1}{2} \\ =&\frac{1}{12f_{2n-1} f_{2n} }+\frac{1}{12f_{2n+1} f_{2n+2} } +\frac{f_{2n-1}}{12 f_{2n} } +\frac{ f_{2n-2} }{12 f_{2n-1} }+\frac{f_{2n+1}}{12 f_{2n+2} }+\frac{ f_{2n} }{12 f_{2n+1} }+\frac{a-3}{6}. \end{split} \end{equation} $

(2.4)

所以

$ \begin{equation*} \begin{split} &S\left ( f_{2n} ,f _{2n+1} \right )+S\left ( f_{2n+1} ,f _{2n+2} \right ) -\frac{f_{2n}}{12f_{2n+1} }-\frac{f_{2n+1}}{12f_{2n+2} }- \frac{1}{12f_{2n+1} f_{2n+2}} \\ =& \frac{a-3}{6}- \left (S\left ( f_{2n-2} ,f _{2n-1} \right )+S\left ( f_{2n-1} ,f _{2n} \right ) -\frac{f_{2n-2}}{12f_{2n-1} }-\frac{f_{2n-1}}{12f_{2n} }- \frac{1}{12f_{2n-1} f_{2n}} \right ) \\ =&S\left ( f_{2n-4} ,f _{2n-3} \right )+S\left ( f_{2n-3} ,f _{2n-2} \right ) -\frac{f_{2n-4}}{12f_{2n-3} }-\frac{f_{2n-3}}{12f_{2n-2} }- \frac{1}{12f_{2n-3} f_{2n-2}} \\ =&\cdots \\ =&\zeta \left ( n \right )\times\left (\frac{a-3}{6} \right ) +\left ( -1 \right )^{n }\times\left [ S\left ( f_{0} ,f _{1} \right )+S\left ( f_{1} ,f _{2} \right )- \frac{f_{0} }{12f_{1} } -\frac{f_{1} }{12f_{2} } -\frac{1 }{12 f_{1}f_{2} } \right ] \\ =&\zeta \left ( n \right )\times \left (\frac{a-3}{6} \right ) +\left ( -1 \right )^{n } \times \left (\frac{a-3}{12} \right ). \end{split} \end{equation*} $

引理 2.2   记$ \left \{ f_{n} \right \} $是双周期Fibonacci数列, $ \left \{ l_{n} \right \} $是双周期Lucas数列, 对正整数$ m $有:

$ \begin{equation} \sum\limits_{n=1}^{m}\left ( \frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1} } \right )=\frac{\left (a+b \right )m }{\alpha } +\sum\limits _{n=1}^{\infty } \left ( \frac{\beta ^{2n+2} }{\left ( ab \right )^{n+1} f_{2n+2} } + \frac{a\beta ^{2n+1} }{\left ( ab \right )^{n+1}f_{2n+1} } \right ) +\mathcal{O}\left (\frac{\beta ^{2m} }{\alpha ^{2m} } \right ), \end{equation} $

(2.5)

$ \begin{equation} \sum\limits_{n=1}^{m}\left ( \frac{l_{2n+1}}{ l_{2n+2}} +\frac{l_{2n}}{l_{2n+1} } \right )=\frac{\left (a+b \right )m }{\alpha } +\sum\limits _{n=1}^{\infty } \frac{\left ( \alpha -\beta \right ) }{\alpha }\left ( \frac{a\beta ^{2n+1} }{\left ( ab \right )^{n+1} l_{2n+2} } + \frac{\beta ^{2n} }{\left ( ab \right )^{n}l_{2n+1} } \right ) +\mathcal{O}\left (\frac{\beta ^{2m} }{\alpha ^{2m} } \right ). \end{equation} $

(2.6)

证   在这里我们仅证明(2.5), (2.6)的证明类似. 根据$ \left \{ f_{n} \right \} $的Binet公式(1.5), 有:

$ \begin{equation*} \begin{split} &\sum\limits_{n=m+1}^{\infty } \frac{ 1 }{ \alpha }\left ( \frac{\beta ^{2n+1} }{\left ( ab \right )^{n} f_{2n+2} }\right ) =\sum\limits_{n=m+1}^{\infty } \frac{b\left ( \alpha -\beta \right ) }{\alpha }\cdot \frac{\beta ^{2n+1} }{\left ( \alpha^{2n+2} - \beta ^{2n+2} \right ) } \\ =&\sum\limits_{n=m+1}^{\infty } \frac{b\left ( \alpha -\beta \right ) }{\alpha }\cdot\frac{1 }{\frac{\alpha ^{2n+2} }{\beta ^{2n+1} } \left ( 1-\frac{\beta ^{2n+2} }{\alpha ^{2n+2} } \right ) } , \end{split} \end{equation*} $

又因为当$ \epsilon \to 0 $时, 有$ \frac{1}{1\pm \epsilon }=1+ \mathcal{O} \left ( \epsilon \right ) $, 而$ 1< \alpha $且$ -1< \beta < 0 $, 所以当$ n\to \infty $时, $ \frac{\beta ^{2n} }{\alpha ^{2n} }\to 0 $, 故$ \frac{1}{1-\frac{\beta ^{2n+2} }{\alpha ^{2n+2} }}= 1+\mathcal{O} \left ( \frac{\beta ^{2n+2} }{\alpha ^{2n+2} } \right ) $, 即有

$ \begin{equation*} \begin{split} &\sum\limits_{n=m+1}^{\infty } \frac{b\left ( \alpha -\beta \right ) }{\alpha }\cdot\frac{1 }{\frac{\alpha ^{2n+2} }{\beta ^{2n+1} } \left ( 1-\frac{\beta ^{2n+2} }{\alpha ^{2n+2} } \right ) } =\sum\limits_{n=m+1}^{\infty } \frac{b\left ( \alpha -\beta \right ) }{\alpha }\cdot\left ( \frac{\beta ^{2n+1} }{\alpha ^{2n+2} }\left ( 1+\mathcal{O} \left ( \frac{\beta ^{2n+2} }{\alpha ^{2n+2} } \right ) \right ) \right )\\ =&\sum\limits_{n=m+1}^{\infty }\left ( \frac{b\left ( \alpha -\beta \right ) }{\alpha }\cdot \frac{\beta ^{2n+1} }{\alpha ^{2n+2} } \right ) + \mathcal{O} \left ( \frac{\beta ^{4m} }{\alpha ^{4m} } \right ) =\mathcal{O} \left ( \frac{\beta ^{4m} }{\alpha ^{4m} } \right )+\mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right )=\mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right ). \end{split} \end{equation*} $

同理,

$ \begin{equation*} \begin{split} &\sum\limits_{n=m+1}^{\infty } \frac{ a }{ \alpha }\left ( \frac{\beta ^{2n } }{\left ( ab \right )^{n} f_{2n+1} }\right ) =\sum\limits_{n=m+1}^{\infty } \frac{ a\left ( \alpha -\beta \right ) }{\alpha }\cdot \frac{\beta ^{2n} }{\left ( \alpha^{2n+1} - \beta ^{2n+1} \right ) } \\ =&\sum\limits_{n=m+1}^{\infty } \frac{a\left ( \alpha -\beta \right ) }{\alpha }\cdot\frac{1 }{\frac{\alpha ^{2n+1} }{\beta ^{2n} } \left ( 1-\frac{\beta ^{2n+1} }{\alpha ^{2n+1} } \right ) } =\sum\limits_{n=m+1}^{\infty } \frac{a\left ( \alpha -\beta \right ) }{\alpha }\cdot\left ( \frac{\beta ^{2n} }{\alpha ^{2n+1} }\left ( 1+\mathcal{O} \left ( \frac{\beta ^{2n+1} }{\alpha ^{2n+1} } \right ) \right ) \right )\\ =&\sum\limits_{n=m+1}^{\infty } \left ( \frac{a\left ( \alpha -\beta \right ) }{\alpha }\cdot \frac{\beta ^{2n} }{\alpha ^{2n+1} } \right ) + \mathcal{O} \left ( \frac{\beta ^{4m} }{\alpha ^{4m} } \right ) =\mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right ) . \end{split} \end{equation*} $

所以,

$ \begin{equation*} \begin{split} &\sum\limits_{n=1}^{m}\left ( \frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1} } \right )=\frac{1}{\alpha } \sum\limits _{n=1}^{m}\left ( \frac{\alpha f_{2n+1}}{ f_{2n+2}} +\frac{\alpha f_{2n}}{f_{2n+1} } \right ) \\ =& \frac{\left (a+b \right )m }{\alpha } -\sum\limits _{n=1}^{m} \frac{1}{\alpha }\left ( \frac{\beta ^{2n+1} }{\left ( ab \right )^{n} f_{2n+2} } + \frac{a\beta ^{2n} }{\left ( ab \right )^{n}f_{2n+1} } \right )\\ =& \frac{\left (a+b \right )m }{\alpha } +\sum\limits _{n=1}^{\infty } \left ( \frac{\beta ^{2n+2} }{\left ( ab \right )^{n+1} f_{2n+2} } + \frac{a\beta ^{2n+1} }{\left ( ab \right )^{n+1}f_{2n+1} } \right ) +\mathcal{O}\left (\frac{\beta ^{2m} }{\alpha ^{2m} } \right ). \end{split} \end{equation*} $

引理 2.3   记$ \left \{ f_{n} \right \} $是双周期Fibonacci数列, $ \left \{ l_{n} \right \} $是双周期Lucas数列, 对正整数$ m $有:

$ \begin{equation} \sum\limits_{n=1}^{m}\frac{1}{f_{2n+1} f_{2n+2} }= \sum\limits_{n=1}^{\infty }\frac{ab+4}{l_{4n+3} +a} +\mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right ) , \end{equation} $

(2.7)

$ \begin{equation} \sum\limits_{n=1}^{m}\frac{1}{l_{2n+1}l_{2n+2} }= \sum\limits_{n=1}^{\infty }\frac{1}{l_{4n+3} -a} +\mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right ) . \end{equation} $

(2.8)

证   在这里我们仅证明(2.7), (2.8)的证明类似. 根据$ \left \{ f_{n} \right \} $的Binet公式(1.5), 有:

$ \begin{equation*} f_{2n+1} f_{2n+2} =\frac{a\left ( \alpha ^{4n+3} +\beta ^{4n+3} +\left ( ab \right ) ^{2n+2} \right ) }{\left ( ab \right )^{2n+1} \left ( \alpha -\beta \right ) ^{2} } =\frac{abl_{4n+3} +a^{2} b}{a^{2} b^{2} +4ab}=\frac{l_{4n+3} +a}{ab+4} , \end{equation*} $

且

$ \begin{equation*} \begin{split} &\sum\limits_{n= m+1}^{\infty }\frac{ab+4}{l_{4n+3} +a}=\frac{\left ( ab+4 \right ) }{a} \sum\limits_{n=m+1}^{\infty } \frac{\left ( ab \right ) ^{2n+2} }{\alpha^{4n+3} + \beta^{4n+3} +\left ( ab \right )^{2n+2} } \\ =&\frac{\left ( ab+4 \right ) }{a} \sum\limits_{n=m+1}^{\infty } \frac{1 }{ \left ( \frac{\alpha ^{4n+3} }{\left ( ab \right )^{2n+2} }\left ( 1+\frac{\beta ^{4n+3} }{\alpha ^{4n+3} } +\frac{ \beta ^{2n+2} }{\alpha ^{2n+1} } \right ) \right ) } \\ =&\frac{\left ( ab+4 \right ) }{a} \sum\limits_{n=m+1}^{\infty } \frac{ \beta ^{2n+2} }{\alpha ^{2n+1} }\left ( 1+\mathcal{O}\left ( \frac{ \beta ^{2m} }{\alpha ^{2m} }\right ) \right ) \\ =& \mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right ) . \end{split} \end{equation*} $

所以,

$ \begin{equation*} \begin{split} &\sum\limits_{n= 1}^{m } \frac{1}{f_{2n+1} f_{2n+2} } =\sum\limits_{n= 1}^{m } \frac{ab+4}{l_{4n+3} +a} =\sum\limits_{n= 1}^{\infty }\frac{ab+4}{l_{4n+3} +a}+\mathcal{O} \left ( \frac{\beta ^{2m} }{\alpha ^{2m} } \right ) . \end{split} \end{equation*} $

在这里我们仅证明(1.7), (1.8)的证明类似. 首先由(2.3) 可得

$ \begin{equation*} \begin{split} &\sum\limits_{n=1}^{m} \left [ S\left ( f_{2n-1} ,f_{2n} \right )+ S\left ( f_{2n} ,f_{2n-1} \right )+S\left ( f_{2n+1} ,f_{2n+2} \right )+S\left ( f_{2n+2} ,f_{2n+1} \right ) \right ] \\ =&\sum\limits_{n=1}^{m} \left [ S\left ( f_{2n-1} ,f_{2n} \right )+ S\left ( f_{2n+1} ,f_{2n+2} \right )+S\left ( f_{2n} ,f_{2n+1} \right )+S\left ( f_{2n-2} ,f_{2n-1} \right ) \right ] \\ =& 2\sum\limits_{n=1}^{m} \left [ S\left ( f_{2n} ,f_{2n+1} \right ) + S\left ( f_{2n+1} ,f_{2n+2} \right )\right ]-S\left ( f_{2m+1} ,f_{2m+2} \right )-S\left ( f_{2m} ,f_{2m+1} \right )\\ &+S\left ( f_{1} ,f_{2} \right )+S\left ( f_{0} ,f_{1} \right ). \end{split} \end{equation*} $

又由(2.4) 可得,

$ \begin{equation*} \begin{split} &\sum\limits_{n=1}^{m} \left [ S\left ( f_{2n-1} ,f_{2n} \right )+ S\left ( f_{2n} ,f_{2n-1} \right )+S\left ( f_{2n+1} ,f_{2n+2} \right )+S\left ( f_{2n+2} ,f_{2n+1} \right ) \right ] \\ =&\frac{1}{12} \sum\limits_{n=1}^{m}\left ( \frac{1 }{f_{2n-1}f_{2n}} +\frac{1}{f_{2n+1}f_{2n+2}} +\frac{f_{2n-1}}{f_{2n}}+\frac{f_{2n-2}}{f_{2n-1}}+\frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1}} \right ) +\frac{m\left ( a-3\right ) }{6}\\ =&\frac{1}{6} \sum\limits_{n=1}^{m}\left ( \frac{1 }{f_{2n+1}f_{2n+2}} +\frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1} } \right ) -\frac{1}{12f_{2m+1}f_{2m+2}}+\frac{1}{12f_{1}f_{2}} -\frac{f_{2m+1}}{12f_{2m+2}}\\ &\quad-\frac{f_{2m}}{12f_{2m+1}} +\frac{f_{0}}{12f_{1}}+\frac{f_{1}}{12f_{2}}+\frac{m\left (a-3 \right ) }{6 }. \end{split} \end{equation*} $

因此, 由引理2.1, 引理2.2和引理2.3可得到

$ \begin{equation*} \begin{split} &2\sum\limits_{n=1}^{m} \left [ S\left ( f_{2n} ,f_{2n+1} \right )+ S\left ( f_{2n+1} ,f_{2n+2} \right ) \right ] \\ =& \frac{1}{6} \sum\limits_{n=1}^{m}\left ( \frac{1 }{f_{2n+1}f_{2n+2}} +\frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1} } \right )+S\left ( f_{2m+1} ,f_{2m+2} \right )+ S\left ( f_{2m} ,f_{2m+1} \right ) \\ &\quad -\frac{1}{12f_{2m+1}f_{2m+2}}-\frac{f_{2m+1}}{12f_{2m+2}}-\frac{f_{2m}}{12f_{2m+1}}+\frac{ \left ( 2m-1 \right ) \left ( a-3 \right ) }{12}\\ =&\frac{1}{6} \sum\limits_{n=1}^{m}\left ( \frac{1 }{f_{2n+1}f_{2n+2}} +\frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1} } \right )+\zeta \left ( m \right ) \times \left ( \frac{a-3}{6} \right ) \\ &\quad +\left ( -1 \right ) ^{m} \times \left ( \frac{a-3}{12} \right ) +\frac{ \left ( 2m-1 \right ) \left ( a-3 \right ) }{12}\\ =&\frac{1}{6} \sum\limits_{n=1}^{m}\left ( \frac{1 }{f_{2n+1}f_{2n+2}} +\frac{f_{2n+1}}{f_{2n+2}} +\frac{f_{2n}}{f_{2n+1} } \right ) +\frac{ \left ( a-3 \right )m }{6}\\ =&\frac{m\left ( a\alpha -3\alpha +a+b \right ) }{6\alpha } \\ & +\frac{1}{6} \sum\limits _{n=1}^{\infty }\left ( \frac{\beta ^{2n+2} }{\left ( ab \right )^{n+1} f_{2n+2} } + \frac{a\beta ^{2n+1} }{\left ( ab \right )^{n+1}f_{2n+1} } +\frac{ab+4}{l_{4n+3} +a} \right ) +\mathcal{O} \left ( \frac{\alpha ^{2m} }{\beta ^{2m} } \right ). \end{split} \end{equation*} $

其中, 因$ a\ge 1 $, $ b\ge 1 $故$ -1< \beta < 0 $且$ 1 <\alpha $.