Appendix
The following lemmas are of great importance for the proof of Theorem 3.
We make use of the notations \(F_{d}(z)=F(d,1,1-d,z),\ G(d)=\frac{\varGamma (1-2d)}{(\varGamma (1-d))^{2}}\), \(\tilde{G}(d)=\frac{\varGamma (1-2d)}{\varGamma (d)\varGamma (1-d)}\) and \(\bar{G}(d)=\frac{\varGamma (d)}{\varGamma (1-d)}\).
Lemma 2
It holds for \(d \in {\mathbb {R}}-{\mathbb {Z}}\), \(i \in {\mathbb {Z}}\) that
$$\begin{aligned} \frac{\varGamma (d-i)}{\varGamma (1-d-i)} = \frac{\varGamma (d+i)}{\varGamma (1-d+i)}. \end{aligned}$$
(24)
Lemma 3
Suppose that the conditions of the Theorem 3 holds. Let \({\zeta } \in [0,1)\), \(|\rho _{j}| < 1\), \(\rho _{j} \ne 0\), \({\zeta } \ne \rho _{j}\) and \({\zeta } \ne 1/\rho _{j}\). Then it holds that for \(m \ge 0\)
$$\begin{aligned}&\textit{1.}\ \ \ \sum \limits _{i=0}^{m}C(d,i,\rho _{j}){\zeta }^{i}\\&\quad = G(d)\Biggl [\frac{\rho _{j}^{2p}}{1-\frac{{\zeta }}{\rho _{j}}} \Biggl ( F_{d}(\rho _{j}) -\frac{{\zeta }}{\rho _{j}} \sum _{v=0}^{m-1} {\zeta }^v \left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k}\right) -\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}\Biggr .\Biggr . \\&\qquad \cdot \Biggl .\left( F_{d}(\rho _{j})-\sum _{v=0}^{m-1} \rho _{j}^{v}\left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k}\right) \right) \Biggr )+F_{d}(\rho _{j}) \\&\qquad \cdot \left( \frac{1-\left( {\zeta }\rho _{j}\right) ^{m+1}}{1-{\zeta } \rho _{j}}\right) +{\zeta }\rho _{j}\sum \limits _{v=1}^{m-1} \left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k}\right) {\zeta }^{v}\frac{1-({\zeta } \rho _{j})^{m-v}}{1-{\zeta }\rho _{j}} \\&\qquad -\Biggl .1\Biggr ].\\&\textit{2.}\ \ \ \sum \limits _{i=0}^{m}C(d,-i,\rho _{j}){\zeta }^{i}\\&\quad = G(d)\Biggl [\rho _{j}^{2p}\Biggl (F_{d}(\rho _{j}) \frac{1-\left( {\zeta }\rho _{j}\right) ^{m+1}}{1-{\zeta }\rho _{j}}+ {\zeta }\rho _{j}\sum \limits _{v=1}^{m-1}\left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k}\right) \Biggr .\Biggr . \\&\qquad \cdot \Biggl .{\zeta }^{v}\frac{1-({\zeta }\rho _{j})^{m-v}}{1-{\zeta }\rho _{j}} -1\Biggr )+\frac{1}{1-\frac{{\zeta }}{\rho _{j}}}\left( F_{d}(\rho _{j}) -\frac{{\zeta }}{\rho _{j}} \sum _{v=0}^{m-1} {\zeta }^{v}\right. \\&\qquad \cdot \left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k}\right) -\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1} \Biggl (F_{d}(\rho _{j})\Biggr .\\&\qquad -\Biggl .\left. \sum _{v=0}^{m-1} \rho _{j}^{v} \left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k} \right) \Biggr )\right) +(\rho _{j}^{2p}-1)\sum \limits _{i=0}^{m} \prod \limits _{k=1}^{i}\left( \frac{d+k-1}{k-d}\right) {\zeta }^{i}\Biggr ]. \end{aligned}$$
Proof
1. We get that
$$\begin{aligned} c_1= & {} \sum \limits _{i=0}^{m}C(d,i,\rho _{j}){\zeta }^{i}=\tilde{G}(d) \left( \rho _{j}^{2p}\sum \limits _{i=0}^{m}\frac{\varGamma (d+i)}{\varGamma (1-d+i)} F_{d+i}(\rho _{j}){\zeta }^{i}\right. \\&\quad +\left. \sum \limits _{i=0}^{m}\frac{\varGamma (d+i)}{\varGamma (1-d+i)}F_{d-i} (\rho _{j}){\zeta }^{i}-\sum \limits _{i=0}^{m}\frac{\varGamma (d+i)}{\varGamma (1-d+i)} {\zeta }^{i}\right) \\&=\tilde{G}(d)(\rho _{j}^{2p}c_{11}+c_{12}-c_{13}). \end{aligned}$$
Furthermore,
$$\begin{aligned} c_{13}=\sum \limits _{i=0}^{m}\frac{\varGamma (d+i)}{\varGamma (1-d+i)}{\zeta }^{i} =\bar{G}(d)\sum \limits _{i=0}^{m}\left( \prod \limits _{k=1}^{i}\frac{d+k-1}{k-d}\right) {\zeta }^{i} \end{aligned}$$
and
$$\begin{aligned} c_{11} = \sum \limits _{i=0}^{m}\frac{\varGamma (d+i)}{\varGamma (1-d+i)}F_{d+i} (\rho _{j}){\zeta }^{i}. \end{aligned}$$
Now it holds that
$$\begin{aligned} \frac{\varGamma (d+i)}{\varGamma (1-d+i)}F_{d+i}(\rho _{j}){\zeta }^{i}= & {} \sum _{j=0}^\infty \frac{\varGamma (d+i+j)}{\varGamma (1-d+i+j)} \rho _{j}^{i+j} ({\zeta }/\rho _{j})^{i} \\= & {} \sum _{v=i}^\infty \frac{\varGamma (d+v)}{\varGamma (1-d+v)} \rho _{j}^{v} ({\zeta }/\rho _{j})^{i} \\= & {} \sum _{v=0}^\infty \! \frac{\varGamma (d\!+\!v)}{\varGamma (1\!-\!d\!+\!v)} \rho _{j}^{v} ({\zeta }/\rho _{j})^{i} \!-\!\! \sum _{v=0}^{i-1} \frac{\varGamma (d\!+\!v)}{\varGamma (1\!-\!d\!+\!v)} \rho _{j}^{v} ({\zeta }/\rho _{j})^{i} \\= & {} ({\zeta }/\rho _{j})^{i} \bar{G}(d) \left( F_{d}(\rho _{j}) - \sum _{v=0}^{i-1} \rho ^v \prod _{k=0}^{v-1} \frac{d+k}{1-d+k} \right) . \end{aligned}$$
Consequently
$$\begin{aligned} c_{11}= & {} \bar{G}(d) \left( F_{d}(\rho _{j}) \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}}{1-\frac{{\zeta }}{\rho _{j}}} - \sum _{i=1}^m \sum _{v=0}^{i-1} {\zeta }^{i} \rho _{j}^{v-i} \prod _{k=0}^{v-1} \frac{d+k}{1-d+k} \right) \\= & {} \bar{G}(d) \left( F_{d}(\rho _{j}) \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}}{1-\frac{{\zeta }}{\rho _{j}}} - \frac{{\zeta }}{\rho _{j}} \frac{\varGamma (1-d)}{\varGamma (d)} \sum _{v=0}^{m-1} {\zeta }^v \frac{1- (\frac{{\zeta }}{\rho _{j}})^{m-v}}{1-\frac{{\zeta }}{\rho _{j}}} \frac{\varGamma (d+v)}{\varGamma (1-d+v)} \right) . \end{aligned}$$
Next we consider the quantity \(c_{12}\). We get that
$$\begin{aligned}&\frac{\varGamma (d+i)}{\varGamma (1-d+i)}F_{d-i}(\rho _{j}){\zeta }^{i}\\&\quad = \sum _{l=0}^\infty \frac{\varGamma (d-i+l)}{\varGamma (1-d-i+l)}\rho _{j}^{l-i} (\rho _{j}{\zeta })^{i} = \sum _{v=-i}^\infty \frac{\varGamma (d+v)}{\varGamma (1-d+v)}\rho _{j}^{v} (\rho {\zeta })^{i} \\&\quad =\sum _{v=0}^\infty \frac{\varGamma (d+v)}{\varGamma (1-d+v)}\rho _{j}^{v}(\rho _{j} {\zeta })^{i}+\sum _{v=1}^{i}\frac{\varGamma (d-v)}{\varGamma (1-v-d)}\rho _{j}^{-v}(\rho _{j} {\zeta })^{i} \\&\quad = (\rho _{j}{\zeta })^{i}\bar{G}(d)\left( F_{d}(\rho _{j})+ \sum _{v=1}^{i}\rho _{j}^{-v}\prod _{k=1}^{v}\frac{d+k-1}{k-d} \right) \end{aligned}$$
and obtain that
$$\begin{aligned} c_{12}= & {} \bar{G}(d) \left( F_{d}(\rho _{j})\sum \limits _{i=0}^{m}\left( {\zeta }\rho _{j} \right) ^{i}+\sum \limits _{i=1}^{m}\left( \prod \limits _{k=1}^{i} \frac{d+k-1}{k-d}\right) {\zeta }^{i}\right. \\&\left. +\sum \limits _{i=1}^{m} \sum \limits _{v=1}^{i-1}\left( \prod \limits _{k=1}^{v} \frac{d+k-1}{k-d}\right) \rho _{j}^{i-v}{\zeta }^{i}\right) \\= & {} \bar{G}(d) \left( F_{d}(\rho _{j})\frac{1-\left( {\zeta }\rho _{j} \right) ^{m+1}}{1-{\zeta }\rho _{j}} + \sum \limits _{i=1}^{m}\left( \prod \limits _{k=1}^{i}\frac{d+k-1}{k-d}\right) {\zeta }^{i}\right. \\&+ \left. {\zeta }\rho _{j}\sum \limits _{v=1}^{m-1} \left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) {\zeta }^{v} \frac{1-({\zeta }\rho _{j})^{m-v}}{1-{\zeta }\rho _{j}}\right) . \end{aligned}$$
Therefore,
$$\begin{aligned} c_{1}= & {} G(d)\left( \rho ^{2p}F_{d}(\rho _{j}) \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}}{1-\frac{{\zeta }}{\rho _{j}}} - \rho _{j}^{2p}\frac{{\zeta }}{\rho _{j}}\sum _{v=0}^{m-1}{\zeta }^{v} \frac{1-(\frac{{\zeta }}{\rho _{j}})^{m-v}}{1-\frac{{\zeta }}{\rho _{j}}} \left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k} \right) \right. \\&+ \left. F_{d}(\rho _{j})\frac{1-\left( {\zeta }\rho _{j}\right) ^{m+1}}{1-{\zeta } \rho _{j}}+{\zeta } \rho _{j}\sum \limits _{v=1}^{m-1} \left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k}\right) {\zeta }^{v}\frac{1-({\zeta }\rho _{j})^{m-v}}{1-{\zeta } \rho _{j}}-1\right) \\= & {} G(d)\left( \rho _{j}^{2p}\frac{1}{1-\frac{{\zeta }}{\rho _{j}}}\left( F_{d} (\rho _{j}) -\frac{{\zeta }}{\rho _{j}}\sum _{v=0}^{m-1}{\zeta }^{v}\left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k}\right) -\left( \frac{{\zeta }}{\rho _{j}} \right) ^{m+1}\left( F_{d}(\rho _{j})\right. \right. \right. \\&-\left. \left. \sum _{v=0}^{m-1}\rho _{j}^{v}\left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k}\right) \right) \right) +F_{d}(\rho _{j}) \left( \frac{1-\left( {\zeta }\rho _{j}\right) ^{m+1}}{1-{\zeta } \rho _{j}}\right) \\&\left. +{\zeta }\rho _{j}\sum \limits _{v=1}^{m-1} \left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k}\right) {\zeta }^{v}\frac{1-({\zeta }\rho _{j})^{m-v}}{1-{\zeta } \rho _{j}}-1\right) . \end{aligned}$$
2. It holds that
$$\begin{aligned} d_{1}=\sum \limits _{i=0}^{m}C(d,-i,\rho _{j}){\zeta }^{i} = \tilde{G}(d) (\rho _{j}^{2p}d_{11}+d_{12}-d_{13}), \end{aligned}$$
where
$$\begin{aligned} d_{11}= & {} \sum \limits _{i=0}^{m}\frac{\varGamma (d-i)}{\varGamma (1-d-i)}F_{d-i}(\rho _{j}){\zeta }^{i},\ \ d_{12} = \sum \limits _{i=0}^{m}\frac{\varGamma (d-i)}{\varGamma (1-d-i)}F_{d+i}(\rho _{j}){\zeta }^{i},\\ d_{13}= & {} \sum \limits _{i=0}^{m}\frac{\varGamma (d-i)}{\varGamma (1-d-i)}{\zeta }^{i}. \end{aligned}$$
Applying Lemma 2 we get that \(d_{11} = c_{12}\), \(d_{12} = c_{11}\), and \(d_{13} = c_{13}\). Thus,
$$\begin{aligned} d_{1}= & {} \tilde{G}(d) (\rho _{j}^{2p}d_{11}+d_{12}-d_{13})=G(d)\left( \rho _{j}^{2p} \left( F_{d}(\rho _{j})\frac{1-\left( {\zeta }\rho _{j}\right) ^{m+1}}{1-{\zeta } \rho _{j}}\right. \right. \nonumber \\&+\left. {\zeta }\rho _{j}\sum \limits _{v=1}^{m-1}\left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k}\right) {\zeta }^{v}\frac{1-({\zeta }\rho _{j})^{m-v}}{1-{\zeta } \rho _{j}}-1\right) +\frac{1}{1-\frac{{\zeta }}{\rho _{j}}}\left( F_{d}(\rho _{j}) \right. \\&-\Biggl .\frac{{\zeta }}{\rho _{j}}\sum _{v=0}^{m-1}{\zeta }^{v} \left( \prod \limits _{k=0}^{v-1}\frac{d+k}{1-d+k}\right) -\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}\left( F_{d}(\rho _{j})\right. \\&\left. \left. - \sum _{v=0}^{m-1}\rho _{j}^v\left( \prod \limits _{k=0}^{v-1} \frac{d+k}{1-d+k}\right) \right) \right) \\&+\left. (\rho _{j}^{2p}-1)\sum \limits _{i=0}^{m} \left( \prod \limits _{k=1}^{i}\frac{d+k-1}{k-d}\right) {\zeta }^{i}\right) . \end{aligned}$$
\(\square \)
Corollary 1
Suppose that the conditions of Theorem 3 hold. Let \({\zeta } \in [0,1)\), \(|\rho _{j}| < 1\), \(\rho _{j} \ne 0\), \({\zeta } \ne \rho _{j}\) and \({\zeta } \ne 1/\rho _{j}\). Then it holds that
-
1.
for \(m \ge 0\ \) difference \(\ D_{1}=\sum \nolimits _{i=1}^{m}C(d,i,\rho _{j}){\zeta }^{m-i} -\sum \nolimits _{i=0}^{m}C(d,i,\rho _{j}){\zeta }^{i-m}\ \) is equal to
$$\begin{aligned} D_{1}= & {} G(d)\left( \left( \rho _{j}^{2p}(F_{d}(\rho _{j})-1) \left( \frac{1-\left( \frac{1}{{\zeta }\rho _{j}}\right) ^{m+1}}{1-\frac{1}{{\zeta } \rho _{j}}}-1\right) \right. \right. \\&+F_{d}(\rho _{j})\left( \frac{1-\left( \frac{\rho _{j}}{{\zeta }} \right) ^{m+1}}{1-\frac{\rho _{j}}{{\zeta }}}-1\right) \\&+\left. \sum \limits _{i=1}^{m}\sum \limits _{v=1}^{i-1} \left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) \left( \frac{\rho _{j}^{i-v}}{{\zeta }^{i}}-\frac{\rho _{j}^{2p}}{{\zeta }^{i} \rho ^{i-v}}\right) \right) {\zeta }^{m}-\left( \rho _{j}^{2p}(F_{d}(\rho _{j})-1) \right. \\&\cdot \left( \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}}{1 -\frac{{\zeta }}{\rho _{j}}}-1\right) +F_{d}(\rho _{j})\left( \frac{1-\left( {\zeta } \rho _{j}\right) ^{m+1}}{1-{\zeta }\rho _{j}}-1\right) \\&+\sum \limits _{i=1}^{m}\sum \limits _{v=1}^{i-1}\left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) \\&\cdot \left. \left. \left( \rho _{j}^{i-v}{\zeta }^{i}-\frac{\rho _{j}^{2p}{\zeta }^{i}}{\rho _{j}^{i-v}}\right) \right) {\zeta }^{-m}-\left( (1+\rho _{j}^{2p})F_{d}(\rho _{j}) -1\right) {\zeta }^{-m}\right) . \end{aligned}$$
-
2.
for \(m < 0\ \) difference \(\ D_{2}=\sum \nolimits _{i=m+1}^{-1}C(d,i,\rho _{j}){\zeta }^{i-m} -\sum \nolimits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{m-i}\ \) is equal to
$$\begin{aligned} D_{2}= & {} G(d)\left( \left( \rho _{j}^{2p}F_{d}(\rho _{j}) \frac{1-\left( \frac{\rho _{j}}{{\zeta }}\right) ^{-m}}{1 -\frac{\rho _{j}}{{\zeta }}}+\left( F_{d}(\rho _{j})-1\right) \frac{1-\left( \frac{1}{{\zeta }\rho _{j}}\right) ^{-m}}{1 -\frac{1}{{\zeta }\rho _{j}}}\right. \right. \\&+(\rho _{j}^{2p}-1)\sum \limits _{i=m+1}^{-1}\left( \prod \limits _{k=i}^{-1} \frac{d-k-1}{-k-d}\right) {\zeta }^{i}\\&\left. +\sum \limits _{i=m+1}^{-2} \sum \limits _{v=i+1}^{-1}\left( \prod \limits _{k=v}^{-1}\frac{d-k-1}{-k-d} \right) \Biggl (\frac{\rho _{j}^{2p}{\zeta }^{i}}{\rho _{j}^{i-v}} -\rho _{j}^{i-v}{\zeta }^{i}\Biggr )\right) {\zeta }^{-m}\\&-\left( \rho _{j}^{2p}F_{d}(\rho _{j})\frac{1-\left( {\zeta }\rho _{j} \right) ^{-m}}{1-{\zeta }\rho _{j}}+\left( F_{d}(\rho _{j})-1\right) \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{-m}}{1-\frac{{\zeta }}{\rho _{j}}} \right. \\&+(\rho _{j}^{2p}-1)\sum \limits _{i=m+1}^{-1}\left( \prod \limits _{k=i}^{-1} \frac{d-k-1}{-k-d}\right) {\zeta }^{-i}+\sum \limits _{i=m+1}^{-2} \sum \limits _{v=i+1}^{-1}\left( \prod \limits _{k=v}^{-1}\frac{d-k-1}{-k-d} \right) \\&\cdot \left. \left. \Biggl (\frac{\rho _{j}^{2p}}{\rho _{j}^{i-v}{\zeta }^{i}} -\frac{\rho _{j}^{i-v}}{{\zeta }^{i}}\Biggr )\right) {\zeta }^{m}-\left( (1 +\rho _{j}^{2p})F_{d}(\rho _{j})-1\right) {\zeta }^{-m}\right) . \end{aligned}$$
Proof
1. We get that
$$\begin{aligned} D_{1}= & {} \sum \limits _{i=1}^{m}C(d,i,\rho _{j}){\zeta }^{m-i} -\sum \limits _{i=1}^{m}C(d,i,\rho _{j}){\zeta }^{i-m}-C(d,0, \rho _{j}){\zeta }^{-m} \\= & {} c_{1}{\zeta }^{m}-c_{2}{\zeta }^{-m}-c_{3}{\zeta }^{-m} \end{aligned}$$
with
$$\begin{aligned} c_{3}=C(d,0,\rho _{j})=G(d)\left( (1+\rho _{j}^{2p})F_{d}(\rho _{j})-1\right) \end{aligned}$$
and
$$\begin{aligned} c_{1}=\sum \limits _{i=1}^{m}C(d,i,\rho _{j}){\zeta }^{-i},\ \ c_{2}=\sum \limits _{i=1}^{m}C(d,i,\rho _{j}){\zeta }^{i}. \end{aligned}$$
Applying Lemma 3 we get that
$$\begin{aligned} c_{1}= & {} G(d)\left( \rho _{j}^{2p}(F_{d}(\rho _{j})-1)\left( \frac{1 -\left( \frac{1}{{\zeta }\rho _{j}}\right) ^{m+1}}{1-\frac{1}{{\zeta } \rho _{j}}}-1\right) +F_{d}(\rho _{j})\left( \frac{1-\left( \frac{\rho _{j}}{{\zeta }} \right) ^{m+1}}{1-\frac{\rho _{j}}{{\zeta }}}-1\right) \right. \\&+ \left. \sum \limits _{i=1}^{m}\sum \limits _{v=1}^{i-1} \left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) \left( \frac{\rho _{j}^{i-v}}{{\zeta }^{i}}-\frac{\rho _{j}^{2p}}{{\zeta }^{i} \rho _{j}^{i-v}}\right) \right) . \end{aligned}$$
Result for \(c_{2}\) is obtained by replacing \({\zeta }\) by \(1/{\zeta }\) in \(c_1\).
$$\begin{aligned} c_{2}= & {} G(d)\left( \rho _{j}^{2p}(F_{d}(\rho _{j})-1)\left( \frac{1 -\left( \frac{{\zeta }}{\rho _{j}}\right) ^{m+1}}{1-\frac{{\zeta }}{\rho _{j}}} -1\right) +F_{d}(\rho _{j})\left( \frac{1-\left( \rho _{j}{\zeta }\right) ^{m+1}}{1 -\rho _{j}{\zeta }}-1\right) \right. \\&\left. +\sum \limits _{i=1}^{m}\sum \limits _{v=1}^{i-1}\left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) \left( \rho _{j}^{i-v}{\zeta }^{i} -\frac{\rho _{j}^{2p}{\zeta }^{i}}{\rho _{j}^{i-v}}\right) \right) . \end{aligned}$$
By inserting the expressions of \(c_1\), \(c_2\), and \(c_3\) we get the desired result.
2. It holds that
$$\begin{aligned} D_{2}= & {} \sum \limits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{i-m} -\sum \limits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{m-i}-C(d,0, \rho _{j}){\zeta }^{-m} \\= & {} d_{1}{\zeta }^{-m}-d_{2}{\zeta }^{m}-d_{3}{\zeta }^{-m} \end{aligned}$$
with
$$\begin{aligned} d_{3}=G(d)\left( (1+\rho _{j}^{2})F_{d}(\rho _{j})-1\right) \end{aligned}$$
and
$$\begin{aligned} d_{1}= & {} \sum \limits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{i}=C(d,0,\rho _{j}) +\sum \limits _{i=1}^{-m-1}C(d,i,\rho _{j}){\zeta }^{i},\\ d_{2}= & {} \sum \limits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{-i} =C(d,0,\rho _{j})+\sum \limits _{i=1}^{-m-1}C(d,i,\rho _{j}){\zeta }^{-i}. \end{aligned}$$
The sum \(d_{1}\) can be obtained by applying Lemma 3 and by replacing m to \(-m-1\). Thus we get that
$$\begin{aligned} d_{1}= & {} G(d)\left( \rho _{j}^{2p}F_{d}(\rho _{j})\frac{1-\left( \frac{\rho _{j}}{{\zeta }}\right) ^{-m}}{1-\frac{\rho _{j}}{{\zeta }}} +(F_{d}(\rho _{j})-1)\frac{1-\left( \frac{1}{{\zeta }\rho }\right) ^{-m}}{1 -\frac{1}{{\zeta }\rho _{j}}}\right. \\&+ (\rho _{j}^{2p}-1)\sum \limits _{i=m+1}^{-1}\left( \prod \limits _{k=i}^{-1} \frac{d-k-1}{-k-d}\right) {\zeta }^{i}+\sum \limits _{i=m+1}^{-2} \sum \limits _{v=i+1}^{-1}\left( \prod \limits _{k=v}^{-1}\frac{d-k-1}{-k-d} \right) \\&\cdot \left. \left( \frac{\rho _{j}^{2p}{\zeta }^{i}}{\rho _{j}^{i-v}} -\rho _{j}^{i-v}{\zeta }^{i}\right) \right) . \end{aligned}$$
Result for \(d_{2}\) is obtained by replacing \({\zeta }\) by \(1/{\zeta }\) in \(d_1\).
$$\begin{aligned} d_{2}= & {} G(d)\left( \rho _{j}^{2p}F_{d}(\rho _{j})\frac{1-\left( {\zeta } \rho _{j}\right) ^{-m}}{1-{\zeta }\rho _{j}}+(F_{d}(\rho _{j})-1) \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{-m}}{1-\frac{{\zeta }}{\rho _{j}}} \right. \\&+ \left. (\rho _{j}^{2p}-1)\sum \limits _{i=m+1}^{-1}\left( \prod \limits _{k=i}^{-1} \frac{d-k-1}{-k-d}\right) {\zeta }^{-i}+\sum \limits _{i=m+1}^{-2} \sum \limits _{v=i+1}^{-1}\left( \prod \limits _{k=v}^{-1}\frac{d-k-1}{-k-d} \right) \right. \\&\cdot \left. \left( \frac{\rho _{j}^{2p}}{\rho _{j}^{i-v}{\zeta }^{i}} -\frac{\rho _{j}^{i-v}}{{\zeta }^{i}}\right) \right) . \end{aligned}$$
By inserting the expressions of \(d_1\), \(d_2\), and \(d_3\) we get the desired result. Hereby we proved corollary. \(\square \)
Corollary 2
Assume that the conditions of Theorem 2 are satisfied. Let \({\zeta } \in [0,1)\), \(|\rho _{j}| < 1\), \(\rho _{j} \ne 0\), \({\zeta } \ne \rho _{j}\) and \({\zeta } \ne 1/\rho _{j}\). Then it holds that
$$\begin{aligned} \textit{1.}\ \ \ \sum \limits _{i=0}^{\infty }C(d,i,\rho _{j}){\zeta }^{i}= & {} G(d)\Biggl [\frac{\rho _{j}^{2p}}{1-\frac{{\zeta }}{\rho _{j}}} \left( F_{d}(\rho _{j})-\frac{{\zeta }}{\rho _{j}}F_{d}({\zeta })\right) -F_{d}({\zeta })\Biggr .\\&+\Biggl .\frac{1}{1-{\zeta }\rho _{j}}\left( F_{d}(\rho _{j})+F_{d}({\zeta }) -1\right) \Biggr ].\\ \textit{2.}\ \ \ \sum \limits _{i=0}^{\infty }C(d,-i,\rho _{j}){\zeta }^{i}= & {} G(d)\Biggl [\frac{\rho _{j}^{2p}}{1-{\zeta }\rho }\left( F_{d}(\rho _{j}) +F_{d}({\zeta })-1\right) -F_{d}({\zeta })\Biggr .\\&+\Biggl .\frac{1}{1-\frac{{\zeta }}{\rho _{j}}}\left( F_{d}(\rho _{j}) -\frac{{\zeta }}{\rho _{j}}F_{d}({\zeta })\right) \Biggr ]. \end{aligned}$$
Proof
This result is immediately obtained from Lemma 3 if we consider the limit as m tends to infinity. \(\square \)
Next we give the proof of Theorem 3.
Proof
From Lemma 1 we have that asymptotic variance of the control statistic for EWMA control charts is defined as (13) and the autocovariance function for ARFIMA (p,d,q) process as (5). Therefore,
$$\begin{aligned} \sigma _{e}^{2}= & {} \sigma ^{2}\frac{\lambda }{2-\lambda }\sum \limits _{l=-q}^{q} \sum \limits _{j=1}^{p}\psi (l)\xi _{j}\sum \limits _{k=-\infty }^{\infty }C(d,p+l-k, \rho _{j})\left( 1-\lambda \right) ^{\left| k\right| } \\&=\sigma ^{2}\frac{\lambda }{2-\lambda }\sum \limits _{l=-q}^{q} \sum \limits _{j=1}^{p}\psi (l)\xi _{j}\sum \limits _{k=-\infty }^{\infty }C(d,m-k, \rho _{j}){\zeta }^{\left| k\right| } \\&=\sigma ^{2}\frac{\lambda }{2-\lambda }\sum \limits _{m=p-q}^{p+q} \sum \limits _{j=1}^{p}\psi (l)\xi _{j}\sum \limits _{i=-\infty }^{\infty }C(d,i, \rho _{j}){\zeta }^{\left| m-i\right| }\\&=\sigma ^{2}\frac{\lambda }{2-\lambda } \sum \limits _{m=p-q}^{p+q}\sum \limits _{j=1}^{p}\psi (l)\xi _{j}S_{mj} \end{aligned}$$
Now we have to calculate sum \(S_{mj}\):
$$\begin{aligned} S_{mj}=\sum \limits _{i=-\infty }^{\infty }C(d,i,\rho _{j}){\zeta }^{\left| m-i\right| } =\sum \limits _{i=-\infty }^{m}C(d,i,\rho _{j}){\zeta }^{m-i}+\sum \limits _{i=m+1}^{\infty } C(d,i,\rho _{j}){\zeta }^{i-m} \end{aligned}$$
For \(m\ge 0\)
$$\begin{aligned}&\sum \limits _{i=0}^{\infty }C(d,-i,\rho _{j}){\zeta }^{m+i}+\sum \limits _{i=1}^{m} C(d,i,\rho _{j}){\zeta }^{m-i}+\sum \limits _{i=0}^{\infty }C(d,i,\rho _{j}){\zeta }^{i-m}\\&\quad -\sum \limits _{i=0}^{m}C(d,i,\rho _{j}){\zeta }^{i-m}. \end{aligned}$$
For \(m<0\)
$$\begin{aligned}&\sum \limits _{i=0}^{\infty }C(d,-i,\rho _{j}){\zeta }^{m+i} -\sum \limits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{m-i} +\sum \limits _{i=0}^{\infty }C(d,i,\rho _{j}){\zeta }^{i-m}\\&\quad +\sum \limits _{i=m+1}^{-1}C(d,i,\rho _{j}){\zeta }^{i-m}. \end{aligned}$$
Hence,
$$\begin{aligned} S_{mj}=\sum \limits _{i=-\infty }^{m}C(d,i,\rho _{j}){\zeta }^{m-i} +\sum \limits _{i=m+1}^{\infty }C(d,i,\rho _{j}){\zeta }^{i-m}={\zeta }^{-m}A_{1} +{\zeta }^{m}A_{2}+A_{3}+A_{4} \end{aligned}$$
with
$$\begin{aligned} A_{1}= & {} \sum \limits _{i=0}^{\infty }C(d,i,\rho _{j}){\zeta }^{i}, \\ A_{2}= & {} \sum \limits _{i=0}^{\infty }C(d,-i,\rho _{j}){\zeta }^{i}, \\ A_{3}= & {} \sum \limits _{i=1}^{m}C(d,i,\rho _{j}){\zeta }^{m-i}-\sum \limits _{i=0}^{m} C(d,i,\rho _{j}){\zeta }^{i-m},\\ A_{4}= & {} \sum \limits _{i=m+1}^{-1}C(d,i,\rho _{j}){\zeta }^{i-m} -\sum \limits _{i=m+1}^{0}C(d,i,\rho _{j}){\zeta }^{m-i}. \end{aligned}$$
From Corollary 1 and Corollary 2 we have that
$$\begin{aligned} \sigma _{e}^{2}= & {} \sigma ^{2}\frac{\varGamma (1-2d)}{(\varGamma (1-d))^{2}} \frac{\lambda }{2-\lambda }\sum \limits _{l=-q}^{q}\sum \limits _{j=1}^{p} \psi (l)\xi _{j}\left[ \frac{{\zeta }^{p+l}+\frac{\rho _{j}^{2p}}{{\zeta }^{p+l}}}{1 -\frac{{\zeta }}{\rho _{j}}}\left( F_{d}(\rho _{j})-\frac{{\zeta }}{\rho }F_{d}({\zeta }) \right) \right. \\&\quad +\,\frac{{\zeta }^{p+l}\rho _{j}^{2p}+\frac{1}{{\zeta }^{p+l}}}{1-{\zeta }\rho _{j}} \left( F_{d}(\rho _{j})+F_{d}({\zeta })-1\right) -\left( {\zeta }^{p+l}+\frac{1}{{\zeta }^{p +l}}\right) F_{d}({\zeta })\\&\quad +\,{\zeta }^{p+l} \left\{ \rho _{j}^{2p}(F_{d}(\rho _{j})-1)\left( \frac{1 -\left( \frac{1}{{\zeta }\rho _{j}}\right) ^{p+l+1}}{1-\frac{1}{{\zeta }\rho _{j}}} -1\right) \right. \\&\quad +\Biggl .F_{d}(\rho _{j})\left( \frac{1-\left( \frac{\rho _{j}}{{\zeta }} \right) ^{p+l+1}}{1-\frac{\rho _{j}}{{\zeta }}}-1\right) \\&\quad \left. +\sum \limits _{i=1}^{p+l} \sum \limits _{v=1}^{i-1}\left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) \left( \frac{\rho _{j}^{i-v}}{{\zeta }^{i}}-\frac{\rho _{j}^{2p}}{{\zeta }^{i} \rho _{j}^{i-v}}\right) \right\} \\&\quad -{\zeta }^{-p-l} \left\{ \rho _{j}^{2p}(F_{d}(\rho _{j})-1) \left( \frac{1-\left( \frac{{\zeta }}{\rho _{j}}\right) ^{p+l+1}}{1 -\frac{{\zeta }}{\rho _{j}}}-1\right) \right. \\&\quad +F_{d}(\rho _{j})\left( \frac{1 -\left( {\zeta }\rho _{j}\right) ^{p+l+1}}{1-{\zeta }\rho _{j}}-1\right) \Biggr .\\&\quad +\left. \Biggl .\sum \limits _{i=1}^{p+l}\sum \limits _{v=1}^{i-1} \left( \prod \limits _{k=1}^{v}\frac{d+k-1}{k-d}\right) \left( \rho _{j}^{i-v}{\zeta }^{i}-\frac{\rho _{j}^{2p}{\zeta }^{i}}{\rho _{j}^{i-v}} \right) \right\} + {\zeta }^{-p-l} \left\{ \rho _{j}^{2p}F_{d}(\rho _{j}) \right. \\&\quad \cdot \frac{1-\left( \frac{\rho _{j}}{{\zeta }}\right) ^{-p-l}}{1 -\frac{\rho _{j}}{{\zeta }}}+(F_{d}(\rho _{j})-1)\frac{1-\left( \frac{1}{{\zeta }\rho _{j}} \right) ^{-p-l}}{1-\frac{1}{{\zeta }\rho _{j}}}+(\rho _{j}^{2p}-1)\\&\quad \cdot \sum \limits _{i=p+l+1}^{-1}\left( \prod \limits _{k=i}^{-1} \frac{d-k-1}{-k-d}\right) {\zeta }^{i}\\&\quad +\left. \sum \limits _{i=p+l+1}^{-2} \sum \limits _{v=i+1}^{-1}\left( \prod \limits _{k=v}^{-1}\frac{d-k-1}{-k-d} \right) \left( \frac{\rho _{j}^{2p}{\zeta }^{i}}{\rho _{j}^{i-v}} -\rho _{j}^{i-v}{\zeta }^{i}\right) \right\} \\&\quad -{\zeta }^{p+l} \left\{ \rho _{j}^{2p}F_{d}(\rho _{j})\frac{1-\left( {\zeta } \rho _{j}\right) ^{-p-l}}{1-{\zeta }\rho _{j}}+(F_{d}(\rho _{j})-1)\frac{1 -\left( \frac{{\zeta }}{\rho _{j}}\right) ^{-p-l}}{1-\frac{{\zeta }}{\rho _{j}}}\right. \end{aligned}$$
$$\begin{aligned}&\qquad +\,(\rho _{j}^{2p}-1)\sum \limits _{i=p+l+1}^{-1}\left( \prod \limits _{k=i}^{-1} \frac{d-k-1}{-k-d}\right) {\zeta }^{-i}+\sum \limits _{i=p+l+1}^{-2} \sum \limits _{v=i+1}^{-1}\left( \prod \limits _{k=v}^{-1}\frac{d-k-1}{-k-d}\right) \\&\qquad \cdot \left. \left. \left( \frac{\rho _{j}^{2p}}{\rho _{j}^{i-v}{\zeta }^{i}} -\frac{\rho _{j}^{i-v}}{{\zeta }^{i}}\right) \right\} -2((1+\rho _{j}^{2p})F_{d} (\rho _{j})-1){\zeta }^{-p-l}\right] . \end{aligned}$$
Thus, we proved the theorem. \(\square \)
In the discussion below the proof of Proposition 1 is given.
Proof
While proving we make use of the notation \(c=(1-d)[(1-d)(1+\beta ^2)-2\beta d]\).
The autocovariance function of ARFIMA(0,d,1) process is given in (8). Hence,
$$\begin{aligned} \sigma _{e}^{2}= & {} \sigma ^{2}\frac{\lambda }{2-\lambda }\tilde{G}(d) \sum \limits _{i=-\infty }^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)} \frac{(1-\beta )^{2}i^{2}-c}{i^{2}-(1-d)^{2}}(1-\lambda )^{\left| i\right| }\\= & {} \sigma ^{2}\frac{\lambda }{2-\lambda }\tilde{G}(d)\left[ \sum \limits _{i=0}^{\infty } \frac{\varGamma (d-i)}{\varGamma (1-d-i)}\frac{(1-\beta )^{2}i^{2}-c}{i^{2}-(1-d)^{2}} (1-\lambda )^{i}\right. \\&\cdot \left. \frac{(1-\beta )^{2}i^{2}-c}{i^{2}-(1-d)^{2}}(1-\lambda )^{i} +\sum \limits _{i=0}^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)}\frac{(1 -\beta )^{2}i^{2}-c}{i^{2}-(1-d)^{2}}(1-\lambda )^{i}\right. \\&-\left. \bar{G}(d)\frac{c}{(1-d)^{2}}\right] =\sigma ^{2}\frac{\lambda }{2-\lambda } \tilde{G}(d)\left[ S_{1}+S_{2}-\bar{G}(d)\frac{c}{(1-d)^{2}}\right] . \end{aligned}$$
From Lemma 2 we have that \(S_{1}=S_{2}\). Now the sum \(S_{2}\) is calculated.
$$\begin{aligned} S_{1}= & {} \sum \limits _{i=0}^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)} \frac{(1-\beta )^{2}i^{2}-c}{i^{2}-(1-d)^{2}}(1-\lambda )^{i} \\= & {} (1-\beta )^{2}\sum \limits _{i=0}^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)} \frac{i^{2}}{i^{2}-(1-d)^{2}}(1-\lambda )^{i} \\&\quad -c\sum \limits _{i=0}^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)} \frac{(1-\lambda )^{i}}{i^{2}-(1-d)^{2}}=(1-\beta )^{2}\sum \limits _{i=0}^{\infty } \frac{\varGamma (d+i)}{\varGamma (1-d+i)}(1-\lambda )^{i}\\&\quad +\,(1-\beta )^{2}(1-d)^{2}\sum \limits _{i=0}^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)} \frac{(1-\lambda )^{i}}{i^{2}-(1-d)^{2}}\\&\quad -c\sum \limits _{i=0}^{\infty }\frac{\varGamma (d+i)}{\varGamma (1-d+i)} \frac{(1-\lambda )^{i}}{i^{2}-(1-d)^{2}}=(1-\beta )^{2}\sum \limits _{i=0}^{\infty } \frac{\varGamma (d+i)}{\varGamma (1-d+i)}(1-\lambda )^{i}\\&\quad +\,[(1-\beta )^{2}(1-d)^{2}-c]\sum \limits _{i=0}^{\infty } \frac{\varGamma (d+i)}{\varGamma (1-d+i)}\frac{(1-\lambda )^{i}}{i^{2}-(1-d)^{2}}. \end{aligned}$$
Using properties of the Gaussian hypergeometric function we get that
$$\begin{aligned} S_{1}=S_{2}=\frac{\varGamma (d)}{\varGamma (1-d)}\left[ (1-\beta )^{2}F_{d}(1-\lambda ) -\frac{(1-\beta )^{2}(1-d)^{2}-c}{(1-d)^{2}}F_{d-1}(1-\lambda )\right] . \end{aligned}$$
Therefore, the asymptotic variance of the control statistic of the EWMA control charts for ARFIMA(0,d,1) process is
$$\begin{aligned} \sigma ^{2}_{e}= & {} \sigma ^{2}\frac{\lambda }{2-\lambda } \frac{\varGamma (1-2d)}{(\varGamma (1-d))^{2}}\left( 2(1-\beta )^{2}F_d(1-\lambda ) -4\beta \frac{2d-1}{1-d}F_{d-1}(1-\lambda )-1-\beta ^{2}\right. \\&+\left. \frac{2\beta d}{1-d}\right) . \end{aligned}$$
\(\square \)
