January 01, 1970
We apply a seemingly forgotten series expression of Nörlund for the psi function to express infinite series involving inverse factorials in closed form. Many of such series contain products of Catalan numbers and (odd) harmonic numbers. We also prove some new series for \(\pi\).
Series associated with a forgotten
.3cm
identity of Nörlund
Kunle Adegoke
Department of Physics and Engineering Physics,
Obafemi Awolowo University, 220005 Ile-Ife
Nigeria
adegoke00@gmail.com
in
Robert Frontczak
Independent Researcher, 72762 Reutlingen
Germany
robert.frontczak@web.de
.2 in
Mathematics Subject Classification: 30B50, 33E20.
Keywords: Series, Psi function, Harmonic number, Catalan number.
The Gamma function, \(\Gamma(z)\), is defined for \(\Re(z)>0\) by the integral [1] \[\Gamma (z) = \int_0^\infty e^{- t} t^{z - 1}\,dt.\] The function \(\Gamma(z)\) has the property that it extends the classical factorial function to the
complex plane by \((z-1)!=\Gamma(z)\). It can be extended to the whole complex plane by analytic continuation. Closely related to the Gamma function is the psi (or digamma) function defined by \(\psi(z)=\Gamma'(z)/\Gamma(z)\). It possesses a range of integral representations and the infinite series expressions. Two of such series expressions are [1] \[\label{psi95expr1}
\psi (z) = - \gamma + \sum_{k=0}^\infty \left ( \frac{1}{k+1}-\frac{1}{k+z} \right ),\tag{1}\] and \[\label{psi95expr2}
\psi (z+n) = \psi(z) + \sum_{k=1}^n \frac{1}{z+k-1},\tag{2}\] where in 1 \(\gamma\) denotes the Euler-Mascheroni constant. Coffey [2] has obtained a variety of other series and integral representations of \(\psi(z)\). Summations over digamma and polygamma functions were studied, among others, by
Alzer et al. [3] and Coffey [4].
A series expression that seems to be not well known is the following expression \[\label{psi95expr3}
\psi (z+h) - \psi(z) = \sum_{n=0}^\infty \frac{(-1)^n\,h\,(h-1)\cdots (h-n)}{(n+1)\,z\,(z+1)\cdots (z+n)},\tag{3}\] which holds for all \(z,h\in\mathbb{C}\) with \(\Re(z)>0\) and \(\Re(z+h)>0\). Identity 3 is stated by Hille in his paper [5] from 1927 and is attributed to Nörlund.
The seemingly forgotten identity 3 will play the key role in this paper. The common feature of all series studied here is the appearance of inverse factorials. Similar series were studied by Sofo [6], [7] and more recently by Karp and Prilepkina [8], and also by Boyadzhiev in his papers from 2020 and 2023 [9], [10]. A finite variant was the subject of another paper by Sofo [11].
Among other things we will establish connections with harmonic numbers \(H_\alpha\) and odd harmonic numbers \(O_\alpha\) defined for \(0\ne\alpha\in\mathbb{C}\setminus\mathbb{Z}^{-}\) by the recurrence relations \[H_\alpha = H_{\alpha - 1} + \frac{1}{\alpha} \qquad \text{and} \qquad O_\alpha = O_{\alpha - 1} + \frac{1}{2\alpha -
1},\] with \(H_0=0\) and \(O_0=0\). Harmonic numbers are connected to the digamma function through the fundamental relation \[\label{Harm95psi}
H_\alpha = \psi(\alpha + 1) + \gamma.\tag{4}\]
Generalized harmonic numbers \(H_\alpha^{(m)}\) and generalized odd harmonic numbers \(O_\alpha^{(m)}\) of order \(m\in\mathbb{C}\) are defined by \[H_\alpha^{(m)} = H_{\alpha - 1}^{(m)} + \frac{1}{\alpha^m} \qquad \text{and} \qquad O_\alpha^{(m)} = O_{\alpha - 1}^{(m)} + \frac{1}{(2\alpha - 1)^m},\] with \(H_0^{(m)}=0\) and \(O_0^{(m)}=0\) so that \(H_\alpha=H_\alpha^{(1)}\) and \(O_\alpha=O_\alpha^{(1)}\).
The recurrence relations imply that if \(\alpha=n\) is a non-negative integer, then \[H_n^{(m)} = \sum_{j = 1}^n \frac{1}{j^m} \qquad \text{and} \qquad O_n^{(m)} = \sum_{j = 1}^n \frac{1}{(2j - 1)^m}.\]
Generalized harmonic numbers are linked to the polygamma functions \(\psi^{(r)}(z)\) of order \(r\) defined by \[\psi^{(r)} (z) = \frac{d^r}{dz^r}\psi(z) = (- 1)^{r + 1} r!\sum_{j = 0}^\infty \frac{1}{{( j + z )^{r + 1} }},\] through \[H_z^{(r)} = \zeta (r) + \frac{(- 1)^{r - 1}}{(r - 1)!} \psi^{(r - 1)} (z + 1),\] where \(\zeta(y)\) is the Riemann zeta function.
To whet the reader’s appetite for reading on, we present the following samples from our results: \[1 + \sum_{n=1}^\infty \frac{1}{n+1} \prod_{j=1}^n \frac{3j-1}{3j+1} = \frac{\pi}{\sqrt{3}},\] \[1 + \sum_{n=1}^\infty \frac{2^n}{n+1} \prod_{j=1}^n \frac{5j-2}{10j+3} = \frac{3\pi}{4}\sqrt{\frac{\sqrt{5}}{\alpha^3}}, \qquad (\alpha = (1+\sqrt{5})/2)\] \[\sum_{n = 0}^\infty \frac{2^{2n}}{\binom{{n}}{h}\binom{{2\left( {n + 1} \right)}}{{\left( {n + 1} \right)}}\left( {n + 1} \right)^2} = \frac{\pi}{4}\,\frac{H_{h - 1/2} - H_{- 1/2}}{\sin (\pi h)},\quad h\in\mathbb{C}\setminus\mathbb{Z},\; 2h\not\in\mathbb{Z}^{-},\] and for non-negative integer \(h\), \[\begin{align} \sum_{n = h}^\infty \frac{{\binom{{2\left( {n - h} \right)}}{{n - h}}}}{{\binom{{n}}{h}\binom{{2\left( {n + 1} \right)}}{{n + 1}}}}\frac{{O_{n - h} }}{{\left( {n + 1} \right)^2 }} &= - \left( { - 1} \right)^h \frac{{\zeta \left( 2 \right) - H_h^{(2)} }}{{2\left( {h + 1} \right)\binom{{2\left( {h + 1} \right)}}{{h + 1}}}} + \left( { - 1} \right)^h \frac{{H_h + 2\ln 2}}{{\left( {h + 1} \right)\binom{{2\left( {h + 1} \right)}}{{h + 1}}}}{O_{h + 1}}\\ &\qquad\qquad - \sum_{n = 0}^{h - 1} {\frac{{\left( { - 1} \right)^{n - h} \binom{{h}}{n}}}{{\binom{{2\left( {h - n} \right)}}{{h - n}}\binom{{2\left( {n + 1} \right)}}{{n + 1}}}}\frac{{O_{n - h} }}{{\left( {n + 1} \right)^2 }}}. \end{align}\]
We firstly prove two results that are an immediate consequence of 3 and which yield some possibly new series for \(\pi\). Two such series provide unexpected expressions involving the golden section \(\alpha=(1+\sqrt{5})/2\). We then evaluate several types of series involving the ratio of Catalan numbers and binomial coefficients in closed form.
Theorem 1. For each integer \(k\geq 2\) we have the expression \[(k-2) \left (1 + \sum_{n=1}^\infty \frac{1}{n+1} \prod_{j=1}^n \frac{kj - (k-2)}{kj+1}\right ) = \pi \cot\left (\frac{\pi}{k}\right ).\] In particular, we have the curious expressions for \(\pi\) in the form \[\label{pi95id1} 1 + \sum_{n=1}^\infty \frac{1}{n+1} \prod_{j=1}^n \frac{3j-1}{3j+1} = \frac{\pi}{\sqrt{3}},\qquad{(1)}\] \[\label{pi95id2} 1 + \sum_{n=1}^\infty \frac{2^n}{n+1} \prod_{j=1}^n \frac{2j-1}{4j+1} = \frac{\pi}{2},\qquad{(2)}\] \[\label{pi95id3} 1 + \sum_{n=1}^\infty \frac{1}{n+1} \prod_{j=1}^n \frac{5j-3}{5j+1} = \frac{\pi}{3} \sqrt{\frac{\alpha^3}{\sqrt{5}}},\qquad{(3)}\] and \[\label{pi95id4} 1 + \sum_{n=1}^\infty \frac{2^n}{n+1} \prod_{j=1}^n \frac{3j-2}{6j+1} = \frac{\sqrt{3} \pi}{4}.\qquad{(4)}\]
Proof. Set \(h=1-2z\) in 3 . Then, for all \(z\in\mathbb{C}\) with \(0<\Re(z)<1\), we have \[\pi \cot (\pi z) = \psi(1-z) - \psi(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \prod_{j=0}^n \left (\frac{1-z}{j+z} - 1\right)\] and with \(z=1/k,k\geq 2,\) the claim follows. ◻
Theorem 2. For all \(z\in\mathbb{C}\) with \(-1/2<\Re(z)<1/2\), we have \[\label{thm295pi95id} \sum_{n=0}^\infty \frac{2^{n+1}}{n+1} \prod_{j=0}^n \frac{2z-j}{2z-(2j+1)} = -\pi \tan (\pi z).\qquad{(5)}\] In particular, when \(z=1/3\) then ?? gives ?? , and when \(z=1/4\) then we get ?? . When \(z=1/5\), then ?? yields \[1 + \sum_{n=1}^\infty \frac{2^n}{n+1} \prod_{j=1}^n \frac{5j-2}{10j+3} = \frac{3\pi}{4}\sqrt{\frac{\sqrt{5}}{\alpha^3}}.\]
Proof. In 3 make the replacements \(z\mapsto 1/2-z\) and \(h\mapsto 2z\), while keeping in mind that \[\psi \left(\frac{1}{2} + z \right ) - \psi \left(\frac{1}{2} - z \right ) = \pi \tan (\pi z).\] ◻
Since \[h (h - 1) \cdots (h - n) = \prod_{j = 0}^n (h - j) = (- 1)^{n + 1} \frac{\Gamma(n + 1 - h)}{\Gamma(- h)}\] and \[z (z + 1) \cdots (z + n) = \prod_{j = 0}^n (z + j) = \frac{\Gamma(n + 1 + z)}{\Gamma(z)},\] identity 3 can be written in terms of harmonic numbers via 4 as \[\label{psi95expr4} H_{z + h - 1} - H_{z - 1} = - \frac{\Gamma (z)}{\Gamma (- h)} \sum_{n = 0}^\infty \frac{\Gamma (n + 1 - h)}{(n + 1)\Gamma (n + 1 + z)}.\tag{5}\]
Lemma 1. We have \[\begin{gather} H_{k - 1/2} = 2O_k - 2\ln 2, \\ H_{k - 1/2}^{(2)} = - 2\zeta \left( 2 \right) + 4O_k^{(2)}, \\ H_{- 1/2}^{(3)} = - 6\zeta \left( 3 \right), \\ H_{k - 1/2}^{(m + 1)} - H_{- 1/2}^{(m + 1)} = 2^{m + 1} O_k^{(m + 1)}. \end{gather}\]
Lemma 2. We have \[\begin{gather} \left( {r - 1/2} \right)! = \frac{{\sqrt \pi }}{{2^{2r} }}\binom{{2r}}{r}r!,\quad r\in\mathbb{C}\setminus\mathbb{Z}^{-},\\ \left( { - r - 1/2} \right)! = \sqrt \pi \frac{{\left( { - 1} \right)^r }}{{r!}}\frac{{2^{2r} }}{{\binom{{2r}}{r}}},\quad r\in\mathbb{N}_0,\\ \end{gather}\]
Theorem 3. We have \[\sum_{n = 0}^\infty \frac{{C_n }}{{2^{2n} \left( {n + 1} \right)}} = 4\left( {1 - \ln 2} \right),\] \[\sum_{n = 0}^\infty \frac{{C_n }}{{2^{2n} \left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{10}}{3} - 4\ln 2,\] and more generally, if \(0\ne z\in\mathbb{C}\setminus\mathbb{Z}^{-}\), \(2z\not\in\mathbb{Z}^{-}\), then \[\label{eq46n0r1wj4} \sum_{n = 0}^\infty \frac{C_n}{{2^{2n}\binom{{n + z}}{n}}} = -2z \left( {H_{z - 1} - H_{z - 1/2} } \right),\qquad{(6)}\] where here and throughout this paper \(C_j\) are Catalan numbers, defined for \(j\in\mathbb{N}_0\) by \[C_j = \frac{\binom{{2j}}{j}}{j + 1}.\]
Proof. Set \(h=1/2\) in 5 , express as factorials and arrange as \[- \sum_{n = 0}^\infty \frac{{\left( {n - 1/2} \right)!}}{{n!\left( {n + 1} \right)}}\frac{1}{z}\frac{{z!n!}}{{\left( {z + n} \right)!}} = 2\sqrt \pi (H_{z - 1} - H_{z - 1/2}),\] that is \[- \sum_{n = 0}^\infty \frac{{\sqrt \pi }}{{2^{2n} }}\frac{{\binom{{2n}}{n}}}{{n + 1}}\frac{1}{z}\frac{1}{{\binom{{z + n}}{n}}} = 2\sqrt \pi (H_{z - 1} - H_{z - 1/2}),\] and hence ?? . ◻
Theorem 4. We have \[\label{eq46nruaw64} \sum_{n = 0}^\infty \frac{{O_{n + 1}}}{{(n + 1)(2n + 1)}} = \frac{\pi^2}{6},\qquad{(7)}\] \[\label{eq46osdi1i1} \sum_{n = 0}^\infty \frac{{C_n }}{{2^{2n} }}\frac{H_{n + 1}}{n + 1} = 8 - \frac{{2\pi ^2 }}{3},\qquad{(8)}\] and more generally, if \(0\ne z\in\mathbb{C}\setminus\mathbb{Z}^{-}\), \(2z\not\in\mathbb{Z}^{-}\), then \[\label{eq46ue5hg8p} \sum_{n = 0}^\infty \frac{{C_n }}{{2^{2n} }}\frac{H_{n + z}}{\binom{{n + z}}{n}} = 2 (H_{z - 1} - H_{z - 1/2})(1 - z H_z) + 2z \left(- H_{z - 1}^{(2)} + H_{z - 1/2}^{(2)} \right).\qquad{(9)}\]
Proof. Differentiate ?? with respect to \(z\) to obtain \[\sum_{n = 0}^\infty \frac{{C_n }}{{2^{2n} }}\frac{{\left( {H_{n + z} - H_z } \right)}}{{\binom{{n + z}}{n}}} = 2H_{z - 1} - 2H_{z - 1/2} + 2z \left( - H_{z - 1}^{(2)} + H_{z - 1/2}^{(2)} \right);\] from which ?? follows upon a second use of ?? . Evaluation of ?? at \(z=1/2\) gives ?? while evaluation at \(z=1\) gives ?? . ◻
Theorem 5. We have \[\sum_{n = 0}^\infty \frac{2n+1}{n+1} \frac{C_n}{2^{2n}} = 4\ln 2,\] \[\sum_{n = 0}^\infty \frac{2n+1}{(n+1)(n+2)} \frac{C_n}{2^{2n}} = 4\ln 2 - 2,\] and more generally, if \(z\in\mathbb{C}\) with \(z - 3/2 \not\in\mathbb{Z}^{-}\), then \[\label{eq46cat95series3} \sum_{n = 0}^\infty \frac{2n+1}{2^{2n}} \frac{C_n}{\binom{{n + z}}{n}} = 2z \left( H_{z - 1} - H_{z - 3/2} \right).\qquad{(10)}\]
Proof. The proof is very similar to the proof of Theorem 3. Set \(h=-1/2\) in 3 and use \(\Gamma(1/2)=\sqrt{\pi}\). ◻
Corollary 6. If \(z\in\mathbb{C}\) with \(z - 3/2\not\in\mathbb{Z}^{-}\), then \[\label{eq46cat95series4} \sum_{n = 0}^\infty \frac{n}{2^{2n}} \frac{C_n}{\binom{{n + z}}{n}} = 2z H_{z - 1} - z\left( H_{z - 1/2} + H_{z - 3/2} \right).\qquad{(11)}\]
Remark 1. We have \[\sum_{n = 0}^\infty \frac{n}{(n+1)} \frac{C_n}{2^{2n}} = \sum_{n = 0}^\infty \frac{2n+1}{(n+1)(n+2)} \frac{C_n}{2^{2n}} = 4\ln 2 - 2.\]
Theorem 7. If \(z\in\mathbb{C}\) with \(z - 3/2\not\in\mathbb{Z}^{-}\), then \[\label{eq46cat95series5} \sum_{n = 0}^\infty \frac{2n+1}{2^{2n}} \frac{C_n H_{n+z}}{\binom{{n + z}}{n}} = 2(z H_z - 1) \left( H_{z - 1} - H_{z - 3/2} \right) + 2z \left( H_{z - 1}^{(2)} - H_{z - 3/2}^{(2)} \right).\qquad{(12)}\] In particular, \[\sum_{n = 0}^\infty \frac{2n+1}{n+1} \frac{C_n H_{n+1}}{2^{2n}} = \frac{2 \pi^2}{3},\] \[\sum_{n = 0}^\infty \frac{2n+1}{(n+2)(n+1)} \frac{C_n H_{n+2}}{2^{2n}} = \frac{2 \pi^2}{3} + 4 \ln 2 - 8,\] and \[\label{eq46nr7v570} \sum_{n = 0}^\infty \frac{O_{n + 2}}{(n + 1)(2n + 3)} = 4 - 2\ln 2 - \frac{\pi^2}{6}.\qquad{(13)}\]
Proof. Differentiate ?? with respect to \(z\) and simplify. Identity ?? is obtained by setting \(z=3/2\) in ?? and using Lemma 1. ◻
Theorem 8. If \(z\in\mathbb{C}\) with \(z - 3/2\not\in\mathbb{Z}^{-}\), then \[\begin{align} \label{eq46cat95series6} \sum_{n = 0}^\infty \frac{n}{2^{2n}} \frac{C_n H_{n+z}}{\binom{{n + z}}{n}} &= (z H_z - 1) \left( 2 H_{z - 1} - H_{z - 1/2} - H_{z - 3/2} \right) \nonumber \\ &\qquad + z \left( 2 H_{z - 1}^{(2)} - H_{z - 1/2}^{(2)} - H_{z - 3/2}^{(2)} \right). \end{align}\qquad{(14)}\] In particular, at \(z=1\) and at \(z=3/2\) we obtain \[\sum_{n = 1}^\infty \frac{n C_n}{2^{2n}}\,\frac{H_{n + 1}}{n + 1} = \frac{2\pi^2}{3} - 4\] and \[\sum_{n = 1}^\infty \frac{n O_{n + 2}}{(n + 1)(2n + 1)(2n + 3)} = \frac{13}{4} - 2\ln 2 - \frac{\pi^2}{6}.\]
Proof. Differentiate ?? with respect to \(z\) and simplify. ◻
Note that we used \[\binom{{n + 3/2}}{n} = \frac{(2n + 1)(2n + 3)}{2^{2n}\,3}\binom{{2n}}{n}.\]
Theorem 9. We have \[\label{eq46pm4exox} \sum_{n = 0}^\infty \frac{{2^{2n} }}{{\binom{{2\left( {n + 1} \right)}}{{n + 1}}\left( {n + 1} \right)^2 }} = \frac{\pi^2}{8},\qquad{(15)}\] \[\label{eq46gujh5lu} \sum_{n = 0}^\infty \frac{1}{{\left( {2n + 1} \right)\left( {n + 1} \right)}} = 2\ln 2,\qquad{(16)}\] and, more generally, for \(h\not\in\mathbb{Z}\), \(2h\not\in\mathbb{Z}^{-}\), \[\label{eq46giggzhv} \sum_{n = 0}^\infty \frac{2^{2n}}{\binom{{n}}{h}\binom{{2\left( {n + 1} \right)}}{{\left( {n + 1} \right)}}\left( {n + 1} \right)^2} = \frac{\pi }{4}\,\frac{H_{h - 1/2} - H_{- 1/2}}{\sin(\pi h)}.\qquad{(17)}\]
Proof. Set \(z=1/2\) in 5 and express as factorials to obtain \[\sum_{n = 0}^\infty \frac{{\left( {n - h} \right)!h!}}{{n!}}\frac{{n!\sqrt \pi}}{{\left( {n + 1/2} \right)!}}\frac{1}{{n + 1}} = h!\left( {- 1 - h} \right)!\left( {H_{- 1/2} - H_{h - 1/2} } \right),\] from which ?? follows upon using Lemma 2 and the Euler reflection formula: \[\left( { - r} \right)!\left( {r - 1} \right)! = \frac{\pi}{\sin(\pi r)},\quad r\not\in\mathbb{Z}.\] Identities ?? and ?? are special cases of ?? at \(h=0\) and \(h=1/2\). Note that in deriving ?? , we used L’Hospital’s rule: \[\begin{align} \lim_{h\to 0}\frac{H_{h - 1/2} - H_{- 1/2}}{\sin(\pi h)} &= \lim_{h\to 0}\frac{{\psi \left( {h + 1/2} \right) - \psi \left( {1/2} \right)}}{\sin (\pi h)} \\ &= \lim_{h\to 0}\frac{\psi^{(1)} \left( {h + 1/2} \right)}{\pi \cos(\pi h)} = \frac{\pi}{2}. \end{align}\] ◻
Corollary 10. If \(h\) is a non-negative integer, then \[\label{eq46t7z1d41} \sum_{n = h}^\infty \frac{{\binom{{2\left( {n - h} \right)}}{{n - h}}}}{{\binom{{n}}{h}\binom{{2\left( {n + 1} \right)}}{{\left( {n + 1} \right)}}\left( {n + 1} \right)^2 }} = (- 1)^h \frac{H_h + 2\ln 2}{\left( {h + 1} \right)\binom{{2\left( {h + 1} \right)}}{{h + 1}}} - \sum_{n = 0}^{h - 1} \frac{(- 1)^{n - h} \binom{{h}}{n}}{\binom{{2\left( {h - n} \right)}}{{\left( {h - n} \right)}}\binom{{2\left( {n + 1} \right)}}{{n + 1}}\left( {n + 1} \right)^2 }.\qquad{(18)}\]
Proof. Write \(h+1/2\) for \(h\) in ?? and use the fact that if \(n\) and \(h\) are non-negative integers, then \[\label{eq46ii21j6e} \binom{{n}}{{h + 1/2}} = \frac{{2^{2n + 2} }}{{\pi \left( {h + 1} \right)}} \begin{cases} \binom{{n}}{h}\binom{{2\left( {n - h} \right)}}{{n - h}}^{ - 1} \binom{{2\left( {h + 1} \right)}}{{h + 1}}^{ - 1},&\text{if n\ge h;} \\ \\ (-1)^{h-n}\binom{{2\left( {h - n} \right)}}{{h - n}}\binom{{h}}{n}^{ - 1} \binom{{2\left( {h + 1} \right)}}{{h + 1}}^{ - 1}&\text{if n\le h;} \\ \end{cases}\tag{6}\] since \[\binom{{n}}{{h + 1/2}} = \begin{cases} \frac{{n!}}{{\left( {n + 1/2} \right)!\left( {n - h - 1/2} \right)!}}\,, &\text{if n\ge h;}\\ \frac{{n!}}{{\left( {n + 1/2} \right)!\left( { - \left( {h - n} \right) - 1/2} \right)!}}\,, &\text{if n\le h.}\\ \end{cases}\] ◻
Theorem 11. If \(h\) is a non-negative integer, then \[\label{last95eq} \begin{align} \sum_{n = h}^\infty {\frac{{\binom{{2\left( {n - h} \right)}}{{n - h}}}}{{\binom{{n}}{h}\binom{{2\left( {n + 1} \right)}}{{n + 1}}}}\frac{{O_{n - h} }}{{\left( {n + 1} \right)^2 }}} &= - \left( { - 1} \right)^h \frac{{\zeta \left( 2 \right) - H_h^{(2)} }}{{2\left( {h + 1} \right)\binom{{2\left( {h + 1} \right)}}{{h + 1}}}} + \left( { - 1} \right)^h \frac{{H_h + 2\ln 2}}{{\left( {h + 1} \right)\binom{{2\left( {h + 1} \right)}}{{h + 1}}}}{O_{h + 1}}\\ &\qquad\qquad - \sum_{n = 0}^{h - 1} {\frac{{\left( { - 1} \right)^{n - h} \binom{{h}}{n}}}{{\binom{{2\left( {h - n} \right)}}{{h - n}}\binom{{2\left( {n + 1} \right)}}{{n + 1}}}}\frac{{O_{h - n} }}{{\left( {n + 1} \right)^2 }}} . \end{align}\qquad{(19)}\]
Proof. Differentiate ?? with respect to \(h\) to obtain \[\sum_{n = 0}^\infty \frac{2^{2n} \left( H_{n - h} - H_h \right)}{\binom{{n}}{h}\binom{{2(n + 1)}}{{n + 1}} (n + 1)^2} = \frac{\pi^2 \cos(\pi h)}{4\sin^2 (\pi h)}\left( H_{h - 1/2} - H_{- 1/2} \right) - \frac{\pi}{4\sin(\pi h)}\left(\zeta (2) - H_{h - 1/2}^{(2)}\right).\] Now write \(h + 1/2\) for \(h\) and use Lemma 1 and identities ?? and 6 . ◻
Note that when \(h=0\) we get \[\sum_{n=0}^\infty \frac{O_n}{(n+1)(2n+1)} = - \frac{\pi^2}{12} + 2 \ln 2,\] which in view of ?? also gives \[\sum_{n=0}^\infty \frac{1}{(n+1)(2n+1)^2} = \frac{\pi^2}{4} - 2 \ln 2.\] When \(h=1\) we see that \[\sum_{n=1}^\infty \frac{O_{n-1}}{(n+1)(2n+1)(2n-1)} = \frac{\pi^2}{36} - \frac{8}{9} \ln 2 + \frac{7}{18}.\]
Theorem 12. If \(m\) is a positive integer, then \[\label{eq46alvbeou} \sum_{n = 0}^\infty \frac{2^{2n}}{(n + 1)\binom{{n + m}}{m}\binom{2(n + m)}{n + m}} = \frac{m}{4}\,\frac{3\,\zeta (2) - 4\,O_{m - 1}^{(2)}}{\binom{2(m - 1)}{m - 1}}.\qquad{(20)}\] In particular, \[\sum_{n = 0}^\infty \frac{2^{2n}}{(n + 1)^2 \binom{2(n + 1)}{n + 1}} = \frac{\pi^2}{8}.\]
Proof. Arrange 5 as \[\sum_{n = 0}^\infty \frac{{\Gamma \left( {n + 1 - h} \right)}}{{\left( {n + 1} \right)\Gamma (n + 1 + z)}} = \Gamma (- h)\left(H_{z - 1} - H_{z + h - 1} \right)\frac{1}{\Gamma (z)}\] and take the limit as \(h\) approaches zero, using \[\lim_{h\to 0}\Gamma (- h)\left( H_{z - 1} - H_{z + h - 1} \right) = \lim_{h\to 0}\Gamma (- h)\left(\psi (z) - \psi (z + h) \right) = \psi^{(1)} (z) = \zeta (2) - H_{z - 1}^{(2)},\] to obtain \[\sum\limits_{n = 0}^\infty {\frac{{n!}}{{\left( {n + 1} \right)\Gamma \left( {n + 1 + z} \right)}}} = \frac{1}{{\Gamma \left( z \right)}}\left( {\zeta \left( 2 \right) - H_{z - 1}^{(2)} } \right).\] Now replace \(z\) with \(z-1/2\) and use \[\label{eq46budvepa} \Gamma \left( {r + 1/2} \right) = \binom{{r - 1/2}}{r}r!\sqrt \pi ,\tag{7}\] to obtain \[\label{eq46z7cdm1p} \sum_{n = 0}^\infty {\frac{1}{{\left( {n + 1} \right)\binom{{n + z}}{z}\binom{{n + z - 1/2}}{{n + z}}}}} = \frac{z}{{\binom{{z - 3/2}}{{z - 1}}}}\left( {\zeta \left( 2 \right) - H_{z - 3/2}^{(2)} } \right),\tag{8}\] which is valid for \(z\in\mathbb{C}\setminus\mathbb{Z}^{-}\) such that \(2z\not\in\mathbb{Z}^{-}\), \(z\ne 0\) and \(z\ne 1/2\). Finally set \(z=m\), a positive integer, in 8 and use \[\label{eq46vgfrd3} \binom{{p - 1/2}}{p} = \frac{{\binom{{2p}}{p}}}{{2^{2p} }};\tag{9}\] which gives ?? on account of Lemma 1. ◻
Theorem 13. If \(m\) is a positive integer, then \[\label{eq46np2sv9u} \begin{align} &\sum_{n = 0}^\infty \frac{2^{2n}O_{n + m}}{(n + 1)\binom{n + m}{m} \binom{{2(n + m)}}{{n + m}}} \\ &\qquad = \frac{m}{4}\,\binom{2(m - 1)}{m - 1}^{-1} \left(7\,\zeta (3) - 8\,O_{m - 1}^{(3)} + O_{m - 1}(3\,\zeta (2) - 4\,O_{m - 1}^{(2)})\right). \end{align}\qquad{(21)}\] In particular, \[\sum_{n = 0}^\infty \frac{2^{2n} O_{n + 1}}{\binom{2(n + 1)}{n + 1}(n + 1)^2} = \frac{7}{4}\,\zeta (3).\]
Proof. Differentiate 8 with respect to \(z\) to obtain \[\begin{align} \sum_{n = 0}^\infty {\frac{{H_{n + z - 1/2} }}{{\left( {n + 1} \right)\binom{{n + z - 1/2}}{{n + z}}\binom{{n + z}}{z}}}} &= \frac{z}{{\binom{{z - 3/2}}{{z - 1}}}}\left( \zeta \left( 2 \right) - H_{z - 3/2}^{(2)} \right)\left( {H_{z - 3/2} - H_{z - 1} } \right) \\ &\qquad + \frac{{2z}}{{\binom{{z - 3/2}}{{z - 1}}}}\left( {\zeta \left( 3 \right) - H_{z - 3/2}^{(3)} } \right) + \frac{{zH_{z - 1} }}{{\binom{{z - 3/2}}{{z - 1}}}}\left( {\zeta \left( 2 \right) - H_{z - 3/2}^{(2)} } \right) \end{align}\] and hence identity ?? on account of Lemma 1 and identity 9 . ◻
Theorem 14. If \(m\) is a positive integer, then \[\label{eq46tu5lpx7} \sum_{n = 0}^\infty {\frac{{C_n}}{{\binom{{n + m}}{n}\binom{{2\left( {n + m} \right)}}{{n + m}}}}} = \frac{m}{2}\,\frac{\left({H_{m - 1} - 2\,O_{m - 1} + 2\,\ln 2}\right)}{{\binom{{2\left( {m - 1} \right)}}{{m - 1}}}}.\qquad{(22)}\]
Proof. Set \(h=1/2\) , write \(z-1/2\) for \(z\) in 5 and use 7 to obtain \[\label{eq46pp9z2ev} \sum_{n = 0}^\infty \frac{{\binom{{n - 1/2}}{n}}}{{\left( {n + 1} \right)\binom{{n + z - 1/2}}{{n + z}}\binom{{n + z}}{z}}} = \frac{{2z}}{{\binom{{z - 3/2}}{{z - 1}}}}\left( H_{z - 1} - H_{z - 3/2} \right),\tag{10}\] from which identity ?? follows ater setting \(z=m\), a positive integer, and using 9 and Lemma 1. ◻
Theorem 15. If \(m\) is a positive integer, then \[\label{eq46h4z23f1} \begin{align} &\sum_{n = 0}^\infty \frac{C_n O_{n + m}}{\binom{{n + m}}{m} \binom{2(n + m)}{n + m}} \\ &\qquad = \frac{m}{4}\,\binom{2(m - 1)}{m - 1}^{- 1} \left( H_{m - 1}^{(2)} + 2\,\zeta (2) - 4\,O_{m - 1}^{(2)} + 2\,O_{m - 1} \left( H_{m - 1} - 2\,O_{m - 1} + 2\ln 2 \right) \right). \end{align}\qquad{(23)}\]
Proof. Differentiate 10 with respect to \(z\), obtaining \[\begin{align} \sum_{n = 0}^\infty {\frac{{\binom{{n - 1/2}}{n}H_{n + z - 1/2} }}{{\left( {n + 1} \right)\binom{{n + z - 1/2}}{{n + z}}\binom{{n + z}}{z}}}} &= \frac{{2z}}{{\binom{{z - 3/2}}{{z - 1}}}}\left( {H_{z - 1}^{(2)} - H_{z - 3/2}^{(2)} - \left( {H_{z - 1} - H_{z - 3/2} } \right)^2 } \right)\\ &\qquad + \frac{{2zH_{z - 1} }}{{\binom{{z - 3/2}}{{z - 1}}}}\left( {H_{z - 1} - H_{z - 3/2} } \right), \end{align}\] from which ?? follows. ◻
Remark 2. When \(m=1\) in Theorem 15 we recover ?? .