Preprint

1 Introduction↩︎

By Stirling’s formula, \(n!\sim \sqrt{2\pi n}(n/e)^n\) as \(n\to+\infty\). Thus, for any integer \(m>1\) we have \[\binom{mk}k\sim\frac{\sqrt m}{\sqrt{2\pi(m-1)k}}\left(\frac{m^m}{(m-1)^{m-1}}\right)^k\] as \(k\to+\infty\). In particular, \[\lim_{k\to+\infty}\root k\of{\binom{4k}k}=\frac{256}{27}.\]

Z.-W. Sun [1] studied series of the type \[\sum_{k=1}^\infty\frac{ak^2+bk+c}{k(3k-1)(3k-2)m^k\binom{4k}k},\] where \(a,b,c\) and \(m\not=0\) are rational numbers. For example, Sun [1] deduced the identities \[\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^{k}}{k(3k-1)(3k-2)\binom{4k}k}=\frac{3}{2}\pi\] and \[\sum_{k=1}^\infty\frac{415k^2-343k+62}{k(3k-1)(3k-2)(-8)^k\binom{4k}k}=-3\log2\] via integration.

Sun’s conjectural identity \[\label{-5} \sum_{k=0}^\infty\frac{(22k^2-92k+11)\binom{4k}{k}}{16^k}=-5\tag{1}\] was confirmed by Max Alekseyev (cf. [2]) via the generating function method. Sun [1] proved further that \[\begin{align} \tag{2}\sum_{k=0}^\infty\frac{(22k^2+17k-2)\binom{4k}k}{(k+1)16^k}&=17, \\\tag{3}\sum_{k=0}^\infty\frac{(11k^2+8k+1)\binom{4k}k}{(3k+1)(3k+2)16^k}&=1, \\\tag{4}\sum_{k=0}^\infty\frac{(22k^2-18k+3)\binom{4k}k}{(2k-1)(4k-1)(4k-3)16^k}&=-\frac{1}{3}. \end{align}\]

Recall that the harmonic numbers are those rational numbers \[H_n=\sum_{0<k\leqslant n}\frac{1}{k}\;\;(n\in\mathbb{N}=\{0,1,2,\ldots\}).\] Sun [1], [3] posed many conjectures on series whose summands involve both \(\binom{4k}k\) and harmonic numbers. In this paper, we aim to prove some of such conjectures.

Our first theorem confirms [1].

Our second theorem confirms the first three identities of [1].

Our third theorem confirms [1].

In the next section we provide some basic lemmas. Theorems 3 -3 will be proved in Sections 3-5 respectively. Our proofs use the basic fact \[\label{Har}H_n=\sum_{k=1}^n\int_0^1t^{k-1}dt=\int_0^1\sum_{k=1}^n t^{k-1}dt =\int_0^1\frac{1-t^n}{1-t}dt\;\;\;(n=1,2,\ldots)\tag{22}\] and the functional equation of the generating function \[f(x)=\sum_{k=0}^{\infty}\binom{4k}{k}x^k\quad\left(|x|<\frac{27}{256}\right).\]

2 Some basic lemmas↩︎

. (i) By (24 ) and (26 ) with \(m=4\), we have \[G_4(x)=1+xG_4(x)^4\;\;\text{and}\;\; f(x)=\frac{G_4(x)}{4-3G_4(x)}=-\frac{1}{3}+\frac{4}{3(4-3G_4(x))}.\label{f61G954}\tag{31}\] Thus \[\sum_{k=0}^{\infty}\frac{\binom{4k}{k}}{3k+1}x^k=G_4(x)=\frac{4f(x)}{3f(x)+1} \label{f441},\tag{32}\] and hence \[\frac{4f(x)}{3f(x)+1}=1+x\left(\frac{4f(x)}{3f(x)+1}\right)^4\] which yields (27 ).

(ii) (28 ) follows from (25 ) and the second equality in 32 .

(iii) By taking derivatives with respect to \(x\), we obtain from (27 ) the identities \[\label{f39} f'(x)=\frac{64f(x)^4}{27f(x)^3-9f(x)-2-256xf(x)^3}\tag{33}\] and \[\label{f3939} f''(x)=\frac{(-81+768x)f(x)^2f'(x)^2+9f'(x)^2+512f(x)^3f'(x)}{27f(x)^3-9f(x)-2-256xf(x)^3}.\tag{34}\] Note that \[27f(x)^3-9f(x)-2-256xf(x)^3=\frac{(3f(x)+1)^2}{f(x)}\] by (27 ). Thus we obtain (29 ) and (30 ) from 33 and 34 .

In view of the above, we have completed the proof of Lemma 2. 0◻

For any positive integer \(j\), we define \[F_j(x)=\sum_{k=0}^\infty\binom{4k}kH_{jk}x^k\quad\text{for}\;|x|<\frac{27}{256}.\]

. In view of 22 , for \(|x|<27/256\) we deduce that \[\label{Fjx} F_j(x)=\sum_{k=1}^\infty\binom{4k}kx^k\int_0^1\frac{1-t^{jk}}{1-t}dt =\int_0^1\frac{f(x)-f(t^jx)}{1-t}dt,\tag{38}\] \[\begin{align} F_j'(x)&=\sum_{k=1}^\infty\binom{4k}kH_{jk}kx^{k-1} =\sum_{k=1}^\infty\binom{4k}kkx^{k-1}\int_0^1\frac{1-t^{jk}}{1-t}dt \\&=\int_0^1\frac{\sum_{k=1}^\infty\binom{4k}kkx^{k-1}-t^j\sum_{k=1}^\infty\binom{4k}kk(t^jx)^{k-1}}{1-t}dt \end{align}\] and thus \[\label{Fj39} F_j'(x)=\int_0^1\frac{f'(x)-t^jf'(t^jx)}{1-t}dt.\tag{39}\] Similarly, for \(|x|<27/256\) we have \[\begin{align} F_j''(x)&=\sum_{k=2}^\infty\binom{4k}kH_{jk}k(k-1)x^{k-2} =\sum_{k=2}^\infty\binom{4k}kk(k-1)x^{k-2}\int_0^1\frac{1-t^{jk}}{1-t}dt \\&=\int_0^1\frac{\sum_{k=2}^\infty\binom{4k}kk(k-1)x^{k-2}-t^{2j}\sum_{k=2}^\infty\binom{4k}kk(k-1)(t^jx)^{k-2}}{1-t}dt \end{align}\] and thus \[\label{Fj3939} F_j''(x)=\int_0^1\frac{f''(x)-t^{2j}f''(t^jx)}{1-t}dt.\tag{40}\]

By (27 ), (29 ) and 30 , for \(0\leqslant t\leqslant 1\) we have \[t^j=\frac{(3f(t^jx)+1)^3(f(t^jx)-1)}{256xf(t^jx)^4}=\varphi(x,f(t^jx)),\] and \[f'(t^jx)=\frac{64f(t^jx)^5}{(3f(t^jx)+1)^2} \;\text{and}\;f''(t^jx)=\frac{4096f(t^jx)^9(9f(t^jx)+5)}{(3f(t^jx)+1)^5}.\] If we set \(y=f(t^jx)\), then \(t^j=\varphi(x,y)\) and \[\frac{dy}{dt}=jf'(t^jx)t^{j-1}x=\frac{64jxy^5}{(3y+1)^2}\varphi(x,y)^{1-1/j}.\] Thus, from 38 we deduce that \[\begin{align} F_j(x)&=\int_{1}^{f(x)}\frac{f(x)-y}{1-\varphi(x,y)^{1/j}}\cdot\frac{(3y+1)^2}{64jxy^5\varphi(x,y)^{1-1/j}}dy\\ &=\int_{1}^{f(x)}\frac{4(y-f(x))(1+3y)^2}{jy((3y+1)^3(y-1)-256xy^4\varphi(x,y)^{1-1/j})}dy. \end{align}\] This proves 35 . Similarly, from 39 we get \[\begin{align} F_j'(x)&= \int_{1}^{f(x)}\frac{f'(x)-\varphi(x,y)\frac{64y^5}{(3y+1)^2}}{1-\varphi(x,y)^{1/j}} \cdot\frac{(3y+1)^2}{64jxy^5\varphi(x,y)^{1-1/j}}dy\\ &=\int_{1}^{f(x)}\frac{4((3y^3-2y^2-y)/(4x)-f'(x))(1+3y)^2}{jy((3y+1)^3(y-1) -256xy^4\varphi(x,y)^{1-1/j})}dy \end{align}\] and thus (36 ) holds. In view of 40 , we also have \[\begin{align} F_j''(x)&= \int_{1}^{f(x)}\frac{f''(x)-\varphi(x,y)^2\frac{4096y^9(9y+5)}{(3y+1)^5}}{1-\varphi(x,y)^{1/j}} \cdot\frac{(3y+1)^2}{64jxy^5\varphi(x,y)^{1-1/j}}dy\\ &=\int_{1}^{f(x)}\frac{4((27y^5-30y^4-16y^3+14y^2+5y)/(16x^2)-f''(x))(1+3y)^2}{jy((3y+1)^3(y-1) -256xy^4\varphi(x,y)^{1-1/j})}dy \end{align}\] and hence (37 ) is true.

In view of the above, we have completed the proof of Lemma 3. 0◻

. Applying 27 with \(x=\frac{1}{16}\), we get \((11\alpha^3-11\alpha^2-7\alpha-1)(\alpha+1)=0\). Since \(\alpha>0\), we see that 41 holds.

By taking the first derivative and the second derivative of \(f(x)\), we see that \[\frac{11}{128}\alpha''-\frac{35}{8}\alpha'+11\alpha=\sum_{k=0}^\infty \frac{(22k^2-92k+11)\binom{4k}{k}}{16^k}.\] Combining this with 1 we obtain 42 . 0◻

3 Proof of Theorem \(\ref{1}\)↩︎

For convenience, for \(j=1,2,3,4\) we set \[A_j=F_j\left(\frac{1}{16}\right),\;A_j'=F_j'\left(\frac{1}{16}\right)\;\text{and}\;A_j''=F_j''\left(\frac{1}{16}\right).\]

. Let \[\sigma_1:=\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{k}-54k+108-\frac{10}{3k}\right).\] Taking the derivative and the second-order derivative of \(F_1(x)\), we deduce that \[\begin{align} \sigma_1 =\frac{22}{256}A_1''-\frac{70}{16}A_1'+11A_1-\frac{54}{16}\alpha'+108(\alpha-1)-\frac{10}{3}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{k16^k}. \end{align}\] Applying 35 –37 with \(j=1\) and \(x=1/{16}\), we get \[A_1=\int_{1}^{\alpha}\frac{4(y-\alpha)(1+3y)^2}{y(y+1)(11y^3-11y^2-7y-1)}dy,\label{F144d}\tag{43}\] \[A_1'=\int_{1}^{\alpha}\frac{4(4(3y^3-2y^2-y)-\alpha')(1+3y)^2}{y(y+1)(11y^3-11y^2-7y-1)}dy, \label{F144dd}\tag{44}\] and \[A_1''=\int_{1}^{\alpha}\frac{4(16(27y^5-30y^4-16y^3+14y^2+5y)-\alpha'')(1+3y)^2}{y(y+1)(11y^3-11y^2-7y-1)}dy.\label{F144ddd}\tag{45}\] So we deduce further that \[\label{sigma1}\sigma_1=\int_{1}^{\alpha}\frac{(27y^2-3y-40)(3y+1)^2}{2y(y+1)}dy -\frac{54}{16}\cdot\frac{64\alpha^5}{(3\alpha+1)^2}+108(\alpha-1)-\frac{40}{3}\log\frac{4\alpha}{3\alpha+1}\tag{46}\] with the aid of the identity \[\begin{align} &\;\;\;(27y^2-3y-40)(11y^3-11y^2-7y-1)\\ &=\frac{11}{16}\left(16(27y^5-30y^4-16y^3+14y^2+5y)-\alpha''\right) \\&\;\;\;-35(4(3y^3-2y^2-y)-\alpha')+11(y-\alpha). \end{align} \label{pqr4414716}\tag{47}\] equivalent to 42 . For the function \[g(y)=\frac{1}{2}(81y^3-54y^2-243y+216)-20\log\frac{2y}{y+1},\] it is easy to verify that \[g'(y)=\frac{(27y^2-3y-40)(3y+1)^2}{2y(y+1)}.\] Thus, from 46 we obtain \[\begin{align} \sigma_1&=g(\alpha)-g(1)-\frac{216\alpha^5}{(3\alpha+1)^2}+108(\alpha-1)-\frac{40}{3}\log\frac{4\alpha}{3\alpha+1}\\ &=\frac{27\alpha(\alpha+1)(11\alpha^3-11\alpha^2-7\alpha-1)}{2(3\alpha+1)^2} \\&\;\;\;-\frac{20}{3}\log{\left(2+\frac{2(5\alpha^2+2\alpha+1)(11\alpha^3-11\alpha^2-7\alpha-1)}{(\alpha+1)^3(3\alpha+1)^2}\right)} \\ &=-\frac{20}{3}\log{2}. \end{align}\] This proves (5 ). 0◻

. Let \[\sigma_2:=\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{2k}+287k-115-\frac{25}{6k}\right)-214+\frac{40}{3}\log{2}.\] Taking the derivative and the second-order derivative of \(F_2(x)\), we deduce that \[\begin{align} \sigma_2 =\frac{22}{256}A_2''-\frac{70}{16}A_2'+11A_2+\frac{287}{16}\alpha'-115(\alpha-1)-\frac{25}{6}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{k16^k}-214+\frac{40}{3}\log{2}. \end{align}\] Applying 35 –37 with \(j=2\) and \(x=\frac{1}{16}\), we get \[A_2=\int_{1}^{\alpha}\frac{2(y-\alpha)}{y((3y+1)(y-1)-4y^2\sqrt{\frac{y-1}{3y+1}})}dy,\] \[A_2'=\int_{1}^{\alpha}\frac{2(4(3y^3-2y^2-y)-\alpha')}{y((3y+1)(y-1)-4y^2\sqrt{\frac{y-1}{3y+1}})}dy,\] \[A_2''=\int_{1}^{\alpha}\frac{2(16(27y^5-30y^4-16y^3+14y^2+5y)-\alpha'')}{y((3y+1)(y-1)-4y^2\sqrt{\frac{y-1}{3y+1}})}dy.\] So we deduce further that \[\label{sigma2} \begin{align} \sigma_2&=I_2+\frac{287}{16}\frac{64\alpha^5}{(3\alpha+1)^2}-115(\alpha-1) -\frac{50}{3}\log\frac{4\alpha}{3\alpha+1}-214+\frac{40}{3}\log{2} \end{align}\tag{48}\] with the aid of the identity 47 , where \[I_2:=\int_{1}^{\alpha}\frac{(27y^2-3y-40)(11y^3-11y^2-7y-1)}{4y((3y+1)(y-1)-4y^2\sqrt{\frac{y-1}{3y+1}})}dy.\]

Putting \(x=1/16\) in (27 ), we get \(\sqrt{\frac{\alpha-1}{3\alpha+1}}=\frac{4\alpha^2}{(3\alpha+1)^2}\). For \(z=\sqrt{(y-1)/{(3y+1)}}\), clearly \[y=-\frac{z^2+1}{3z^2-1}\;\;\text{and}\;\;dy=\frac{8z}{(3z^2-1)^2}dz.\] So \[\begin{align}I_2 &=\int_{0}^{\sqrt{\frac{\alpha-1}{3\alpha+1}}}\frac{16(z^3-z^2+3z+1)(81z^4-75z^2+4)}{(z-1)(3z^2-1)^4 (z^2+1)}dz\\ &=\int_{0}^{\frac{4\alpha^2}{(3\alpha+1)^2}}\frac{16(z^3-z^2+3z+1)(81z^4-75z^2+4)}{(z-1)(3z^2-1)^4 (z^2+1)}dz. \end{align}\]

For the function \[\begin{align} g_2(z)=\frac{4z(486z^5+135z^4-351z^3-126z^2+27z+11)}{(3z^2-1)^3}+10\log\frac{(z-1)^2}{z^2+1}, \end{align}\] it is easy to verify that \[g_2'(z)=\frac{16(z^3-z^2+3z+1)(81z^4-75z^2+4)}{(z-1)(3z^2-1)^4 (z^2+1)}.\] Thus, \[I_2=g_2\left(\frac{4\alpha^2}{(3\alpha+1)^2}\right)-g_2(0).\] Combining this with 48 , we obtain \[\begin{align} \sigma_2&=\frac{(11\alpha^3-11\alpha^2-7\alpha-1)p_1(\alpha)}{(3\alpha+1)^2 (33\alpha^4+108\alpha^3+54\alpha^2+12\alpha+1)^3}+\frac{10}{3}\log\frac{(\alpha+1)^6(3\alpha+1)^5(5\alpha+1)^6}{64\alpha^5(97\alpha^4+ 108 \alpha^3+ 54\alpha^2+12\alpha+1)^3}\\ &=\frac{10}{3}\log\left(1-\frac{(11\alpha^3-11\alpha^2-7\alpha-1)p_2(\alpha)}{64\alpha^5(97\alpha^4+ 108 \alpha^3+ 54\alpha^2+12\alpha+1)^3}\right) =0, \end{align}\] where \[\begin{align} p_1(y)&=3750516 y^{14}+40573764 y^{13}+ 178503183 y^{12}+ 437065524 y^{11}+ 654760746 y^{10}\\ &\;\;\;+649292868 y^9+447600789 y^8+ 221967000 y^7+ 80835264 y^6+ 21786588 y^5 \\ &\;\;\;+4319865 y^4+ 615636 y^3+ 59938 y^2+ 3580 y+99, \end{align}\] and \[\begin{align} p_2(y)&=4964927 y^{14}+ 19641236 y^{13}+ 39021903 y^{12}+ 50600432 y^{11} \\&\;\;\;+ 47083971 y^{10}+ 32761508 y^9 +17347827 y^8+ 7024768 y^7+ 2167757 y^6 \\&\;\;\;+ 504220 y^5+ 86653 y^4+ 10640 y^3+ 881 y^2+ 44 y+1. \end{align}\] This proves (6 ). 0◻

. Let \[\sigma_3:=\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{3k}-296k+178-\frac{25}{3k}\right)+196+\frac{80}{3}\log{2}.\] Taking the derivative and the second-order derivative of \(F_3(x)\), we deduce that \[\begin{align} \sigma_3 =\frac{22}{256}A_3''-\frac{70}{16}A_3'+11A_3-\frac{296}{16}\alpha'+178(\alpha-1)-\frac{25}{3}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{k16^k}+196+\frac{80}{3}\log{2}. \end{align}\]

Applying 35 –37 with \(j=3\) and \(x=1/{16}\), we get \[A_3=\int_{1}^{\alpha}\frac{4(y-\alpha)}{3y(y-1)(3y+1-2\sqrt[3]{2} y\sqrt[3]{\frac{y}{y-1}})}dy, \label{F344d}\tag{49}\] \[A_3'=\int_{1}^{\alpha}\frac{4(4(3y^3-2y^2-y)-\alpha')}{3y(y-1)(3y+1-2\sqrt[3]{2} y\sqrt[3]{\frac{y}{y-1}})}dy,\label{F344dd}\tag{50}\] \[A_3''=\int_{1}^{\alpha}\frac{4(16(27y^5-30y^4-16y^3+14y^2+5y)-\alpha'')}{3y(y-1)(3y+1-2\sqrt[3]{2} y\sqrt[3]{\frac{y}{y-1}})}dy. \label{F344ddd}\tag{51}\] So we deduce further that \[\label{sigma3} \begin{align} \sigma_3&=I_3 -\frac{296}{16}\cdot\frac{64\alpha^5}{(3\alpha+1)^2}+178(\alpha-1) -\frac{100}{3}\log\frac{4\alpha}{3\alpha+1}+196+\frac{80}{3}\log{2} \end{align}\tag{52}\] with the aid of the identity 47 , where \[I_3:=\int_{1}^{\alpha}\frac{(27y^2-3y-40)(11y^3-11y^2-7y-1)}{6y(y-1)(3y+1-2\sqrt[3]{2} y\sqrt[3]{\frac{y}{y-1}})}dy.\]

Putting \(x=1/16\) in (27 ), we get \(\sqrt[3]{\frac{\alpha-1}{\alpha}}=\frac{2\sqrt[3]{2}\alpha}{3\alpha+1}\). For \(z=\sqrt[3]{(y-1)/{y}}\), clearly \[y=\frac{1}{1-z^3}\;\;\text{and}\;\;dy=\frac{3z^2}{(z^3-1)^2}dz.\] So \[\begin{align}I_3 &=\int_{0}^{\sqrt[3]{\frac{\alpha-1}{\alpha}}}\frac{(40z^6-83z^3+16)(z^9-10z^6+28z^3-8)}{2(z^3-1)^4(z^4-4z+2\sqrt[3]{2})}dz\\ &=\int_{0}^{\frac{2\sqrt[3]{2}\alpha}{3\alpha+1}}\frac{(40z^6-83z^3+16)(z^9-10z^6+28z^3-8)}{2(z^3-1)^4(z^4-4z+2\sqrt[3]{2})}dz. \end{align}\]

For the function \[\begin{align} g_3(z)&=\frac{z(72z^8+40\sqrt[3]{2}z^7+20\sqrt[3]{4}z^6-135z^5-80\sqrt[3]{2}z^4-44\sqrt[3]{4}z^3+36z^2+22\sqrt[3]{2}z+12\sqrt[3]{4})}{2(z^3-1)^3}\\ &\;\;\;-\frac{20}{3}\log\frac{2}{(\sqrt[3]{2}-z)^3}, \end{align}\] it is easy to verify that \[g_3'(z)=\frac{(40z^6-83z^3+16)(z^9-10z^6+28z^3-8)}{2(z^3-1)^4(z^4-4z+2\sqrt[3]{2})}.\] Thus, \[I_3=g_3\left(\frac{2\sqrt[3]{2}\alpha}{3\alpha+1}\right)-g_3(0).\] Combining this with 52 , we obtain \[\begin{align} \sigma_3&=-\frac{2(11\alpha^3-11\alpha^2-7\alpha-1)p_3(\alpha)}{(3\alpha+1)^2 (11\alpha^3+27\alpha^2+9\alpha+1)^3} \\&\;\;\;-\frac{20}{3} \log\left(1+\frac{(11\alpha^3-11\alpha^2-7\alpha-1)(5\alpha^2+2\alpha+1)}{(\alpha+1)^3(3\alpha+1)^2}\right)\\ &=0, \end{align}\] where \[\begin{align} p_3(y)&=71632y^{11}+ 599104y^{10}+ 2018295y^9+ 3436765y^8+3356404y^7 +2077344y^6\\ &\;\;\;+ 845834y^5+ 229586y^4+ 41220y^3+ 4712y^2+ 311y+9 . \end{align}\] This proves (7 ). 0◻

. Let \[\sigma_4:=\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{4k}-\frac{449}{2}k+\frac{275}{2}-\frac{85}{12k}\right)+151+\frac{80}{3}\log{2}.\] Taking the derivative and the second-order derivative of \(F_3(x)\), we deduce that \[\begin{align} \sigma_4 =\frac{22}{256}A_4''-\frac{70}{16}A_4'+11A_4-\frac{449}{32}\alpha'+\frac{275}{2}(\alpha-1)-\frac{85}{12}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{k16^k}+151+\frac{80}{3}\log{2}. \end{align}\]

Applying 35 –37 with \(j=4\) and \(x=\frac{1}{16}\), we get \[A_4=\int_{1}^{\alpha}\frac{y-\alpha}{y(y-1)(3y+1-2y\sqrt[4]{\frac{3y+1}{y-1}})}dy,\label{F444d}\tag{53}\] \[A_4'=\int_{1}^{\alpha}\frac{4(3y^3-2y^2-y)-\alpha'}{y(y-1)(3y+1-2y\sqrt[4]{\frac{3y+1}{y-1}})}dy,\label{F444dd}\tag{54}\] \[A_4''=\int_{1}^{\alpha}\frac{16(27y^5-30y^4-16y^3+14y^2+5y)-\alpha''}{y(y-1)(3y+1-2y\sqrt[4]{\frac{3y+1}{y-1}})}dy.\label{F444ddd}\tag{55}\]

So we deduce further that \[\label{sigma4} \begin{align} \sigma_4&=I_4-\frac{449}{32}\cdot\frac{64\alpha^5}{(3\alpha+1)^2}+\frac{275}{2}(\alpha-1) -\frac{85}{3}\log\frac{4\alpha}{3\alpha+1}+151+\frac{80}{3}\log{2} \end{align}\tag{56}\] with the aid of the identity 47 , where \[I_4:=\int_{1}^{\alpha}\frac{(27y^2-3y-40)(11y^3-11y^2-7y-1)}{8y(y-1)(3y+1-2y\sqrt[4]{\frac{3y+1}{y-1}})}dy .\]

Putting \(x=1/16\) in (27 ), we get \(\sqrt[4]{\frac{\alpha-1}{3\alpha+1}}=\frac{2\alpha}{3\alpha+1}\). For \(z=\sqrt[4]{(y-1)/{(3y+1)}}\), clearly \[y=-\frac{z^4+1}{3z^4-1}\;\;\text{and}\;\;dy=\frac{16z^3}{(3z^4-1)^2}dz.\] So \[\begin{align}I_4 &=\int_{0}^{\sqrt[4]{\frac{\alpha-1}{3\alpha+1}}}\frac{8(z^3-z^2+z+1)(z^6-z^4+3z^2+1) (81z^8-75z^4+4)}{(z-1)(3z^4-1)^4(z^4+1)}dz\\ &=\int_{0}^{\frac{2\alpha}{3\alpha+1}}\frac{8(z^3-z^2+z+1)(z^6-z^4+3z^2+1)(81z^8-75z^4+4)}{(z-1) (3z^4-1)^4(z^4+1)}dz. \end{align}\]

For the function \[\begin{align} g_4(z)&=\frac{2zQ(z)}{(3 z^4-1)^3}+5\log\frac{(z-1)^4}{z^4+1} \end{align}\] with \[Q(z)=486z^{11}+270z^{10}+135z^9+54z^8-351z^7-216z^6-126z^5-68z^4+27z^3+18z^2+11z+6,\] it is easy to verify that \[g_4'(z)=\frac{8(z^3-z^2+z+1)(z^6-z^4+3z^2+1) (81z^8-75z^4+4)}{(z-1) (3z^4-1)^4(z^4+1)}.\] Thus, \[I_4=g_4\left(\frac{2\alpha}{3\alpha+1}\right)-g_4(0).\] Combining this with 56 and the identity \[\frac{(3\alpha+1)^{15}}{\alpha^{15}}=\left(\frac{16\alpha}{\alpha-1}\right)^5\] obtained from (27 ) with \(x=1/{16}\), we find that \[\begin{align} \sigma_4&=-\frac{(11\alpha^3-11\alpha^2-7\alpha-1)p_4(\alpha)}{2(1+3\alpha)^2 (11\alpha^3+27\alpha^2+9\alpha+1)^3} \\&\;\;\;+\frac{5}{3}\log\frac{(\alpha+1)^{12}(3\alpha+1)^{17}}{262144\alpha^{17}(97\alpha^4+108\alpha^3+54\alpha^2+ 12\alpha+1)^3}\\ &=\frac{5}{3}\log\frac{16^5\alpha^3(\alpha+1)^{12} (3\alpha+1)^2}{262144(\alpha-1)^5 (97\alpha^4+108\alpha^3+54\alpha^2+ 12\alpha+1)^3}\\ &=\frac{5}{3}\log\left(1-\frac{(11\alpha^3-11\alpha^2-7\alpha-1)p_5(\alpha)}{(\alpha-1)^5 (97\alpha^4+108\alpha^3+54\alpha^2+ 12\alpha+1)^3}\right) =0, \end{align}\] where \[\begin{align} p_4(y)&=5867532y^{14}+63476028y^{13}+276465231y^{12}+604283868y^{11}+773801154y^{10} \\&\;\;\;+655999884y^9+390925197y^8+169175304y^7+54045864y^6+12792132y^5 \\&\;\;\;+2221641y^4+275580y^3+23114y^2+1172y+27 \end{align}\] and \[\begin{align} p_5(y)&=82967 y^{14}- 54788 y^{13}- 111085y^{12}+ 20604 y^{11}+ 108295 y^{10}+ 41488 y^9\\ &\;\;\;-30309y^8- 30728y^7 - 5139y^6 +5660y^5+ 4177y^4+ 1356y^3+ 245y^2+ 24y+1 . \end{align}\] This proves (8 ). 0◻

4 Proof of Theorem \(\ref{2}\)↩︎

. This can be easily proved by induction or the Abel summation. 0◻

. Let \(A(x)=G_m(x)-1,\Phi(x)=(1+x)^m\). Then \(A(x)=x\Phi(A(x))\). Applying Lagrange’s inversion formula (cf.[7]), for each \(k=1,2,3,\ldots\) we have \[k[x^k]G_m(x)^n=k[x^k]H(A(x))=[z^{k-1}]H'(z)\Phi(z)^k=[z^{k-1}]n(1+z)^{mk+n-1},\] where \(H(z)=(1+z)^n\), and \([x^k]\Psi(x)\) denotes the coefficient of \(x^k\) in the power series expansion of \(\Psi(x)\). This proves 59 . 0◻

. Let \(\alpha=f(\frac{1}{16})\). Putting \(x=1/16\) in (32 ), (60 ) and (61 ), we find \[\begin{align} \sum_{k=0}^\infty\frac{\binom{4k}{k}}{(3k+1)16^k}&=\frac{4\alpha}{3\alpha+1}, \\ \sum_{k=0}^\infty\frac{\binom{4k}{k}(8k+2)}{(3k+1)(3k+2)16^k}&=\left(\frac{4\alpha}{3\alpha+1}\right)^2, \\ \sum_{k=0}^\infty\frac{\binom{4k}{k}(4k+1)(4k+2)}{(k+1)(3k+1)(3k+2)16^k}& =\left(\frac{4\alpha}{3\alpha+1}\right)^3. \end{align}\] Thus, with the aid of 29 we deduce that \[\begin{align} &\;\;\;\sum_{k=1}^\infty\frac{\binom{4k}{k}(638k^4+966k^3+194k^2-199k-59)}{(k+1)(3k+1)(3k+2)16^k}\\ &=\sum_{k=1}^\infty\frac{\binom{4k}{k}(4k+1)(4k+2)}{(k+1)(3k+1)(3k+2)16^k}+\frac{23}{18}\sum_{k=1}^\infty\frac{\binom{4k}{k}(8k+2)}{(3k+1)(3k+2)16^k} \\ &\;\;\;+\frac{8}{3}\sum_{k=1}^\infty\frac{\binom{4k}{k}}{(3k+1)16^k} +\frac{638}{9}\sum_{k=1}^\infty\frac{\binom{4k}{k}k}{16^k} -\frac{310}{9}\sum_{k=1}^\infty\frac{\binom{4k}{k}}{16^k}\\ &=\left(\frac{4\alpha}{3\alpha+1}\right)^3-1+\frac{23}{18}\left(\left(\frac{4\alpha}{3\alpha+1}\right)^2-1\right) +\frac{8}{3}\left(\frac{4\alpha}{3\alpha+1}-1\right) \\&\;\;\;+\frac{638}{9\times16}\cdot\frac{64\alpha^5}{(3\alpha+1)^2}-\frac{310}{9}(\alpha-1)\\ &=\frac{2(11\alpha^3-11\alpha^2- 7\alpha-1)(348 \alpha^3+ 464\alpha^2+ 305\alpha+99)}{9(3 \alpha+1)^3}+\frac{103}{2}=\frac{103}{2}. \end{align}\] Similarly, we have \[\begin{align} &\;\;\;\sum_{k=1}^\infty\frac{\binom{4k}{k}(770k^4+1134k^3+82k^2-381k-105)}{(k+1)(3k+1)(3k+2)16^k}\\ &=-\sum_{k=1}^\infty\frac{\binom{4k}{k}(4k+1)(4k+2)}{(k+1)(3k+1)(3k+2)16^k} -\frac{43}{18}\sum_{k=1}^\infty\frac{\binom{4k}{k}(8k+2)}{(3k+1)(3k+2)16^k} \\ &\;\;\;-4\sum_{k=1}^\infty\frac{\binom{4k}{k}}{(3k+1)16^k}+\frac{770}{9} \sum_{k=1}^\infty\frac{\binom{4k}{k}k}{16^k} -\frac{406}{9}\sum_{k=1}^\infty\frac{\binom{4k}{k}}{16^k}\\ &=-\left(\frac{4\alpha}{3\alpha+1}\right)^3+1-\frac{43}{18}\left(\left(\frac{4\alpha}{3\alpha+1}\right)^2-1\right) -4\left(\frac{4\alpha}{3\alpha+1}-1\right) \\&\;\;\;+\frac{770}{9\times16}\cdot\frac{64\alpha^5}{(3\alpha+1)^2}-\frac{406}{9}(\alpha-1)\\ &=\frac{2(11\alpha^3-11\alpha^2- 7\alpha-1)(420 \alpha^3+ 560 \alpha^2+ 329 \alpha+27)}{9(3 \alpha+1)^3}+\frac{117}{2}=\frac{117}{2}. \end{align}\] This concludes the proof. 0◻

. Observe that \[\begin{align} \frac{11k^2+8k+1}{(3k+1)(3k+2)}&=-\frac{22k^2-92k+11}{5}-\frac{3(-176k^3-48k^2+80k+24)}{40(3k+1)} \\&\;\;\;+\frac{3(-176k^3+384k^2+224k+24)}{8(3k+1)(3k+2)}. \end{align}\label{kk}\tag{65}\]

Putting \(m=16\) and \(\psi(k)=H_{2k}\) in (57 ) and (58 ) and letting \(n\to +\infty\), we then obtain \(S_1=0=S_2\), where \(S_1\) denotes the expression \[\sum_{k=1}^{\infty}\frac{\binom{4k}{k}H_{2k}(-176k^3+384k^2+224k+24)}{(3k+1)16^k} +\sum_{k=0}^{\infty}\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}}{(3k+1)16^k}(H_{2k+2}-H_{2k})\] and \(S_2\) denotes the expression \[\sum_{k=1}^{\infty}\frac{\binom{4k}{k}H_{2k}(-176k^3-48k^2+80k+24)}{(3k+1)(3k+2)16^k} +\sum_{k=0}^{\infty}\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}}{(3k+1)(3k+2)16^k}(H_{2k+2}-H_{2k}).\]

In light of 65 and the fact that \(S_1=0=S_2\), we have \[\begin{align} &\;\;\;\frac{1}{5}\sum_{k=1}^\infty\frac{\binom{4k}{k}}{16^k}\left(P(k)H_{2k}+287k-115-\frac{25}{6k}\right) +\sum_{k=1}^\infty \frac{\binom{4k}{k}((11k^2+8k+1)H_{2k}+\frac{23}{2}k+\frac{17}{2}+\frac{5}{3k})}{(3k+1)(3k+2)16^k} \\ &=\frac{1}{5}\sum_{k=1}^\infty\frac{\binom{4k}{k}(287k-115-\frac{25}{6k})}{16^k} +\frac{3}{40}\left(\sum_{k=1}^\infty\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}}{(3k+1)16^k} (H_{2k+2}-H_{2k})+36\right)\\ &\;\;\;-\frac{3}{8}\left(\sum_{k=1}^\infty\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}}{(3k+1)(3k+2)16^k} (H_{2k+2}-H_{2k})+18\right)+\sum_{k=1}^\infty\frac{\binom{4k}{k}\left(\frac{23}{2}k+\frac{17}{2}+\frac{5}{3k}\right)}{(3k+1)(3k+2)16^k} \\ &=\frac{9}{10}\sum_{k=1}^\infty\frac{\binom{4k}{k}(638k^4+966k^3+194k^2-199k-59)}{(k+1)(3k+1)(3k+2)16^k} -\frac{81}{20}=\frac{423}{10} \end{align}\] with the aid of 62 . Thus, we have \[\begin{align} &\;\sum_{k=1}^\infty \frac{\binom{4k}{k}((11k^2+8k+1)H_{2k}+\frac{23}{2}k+\frac{17}{2}+\frac{5}{3k})}{(3k+1)(3k+2)16^k} \\=&\;-\frac{1}{5}\sum_{k=1}^\infty\frac{\binom{4k}{k}}{16^k}\left(P(k)H_{2k}+287k-115-\frac{25}{6k}\right)+\frac{423}{10}. \end{align}\] So 64 follows from 6 .0◻

Proof. Putting \(m=16\) and \(\psi(k)=H_{4k}\) in 57 and 58 , and letting \(n\to +\infty\), we then obtain \(S_3=0=S_4\), where \(S_3\) denotes the expression \[\sum_{k=1}^{\infty}\frac{\binom{4k}{k}H_{4k}(-176k^3+384k^2+224k+24)}{(3k+1)16^k} +\sum_{k=0}^{\infty}\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}(H_{4k+4}-H_{4k})}{(3k+1)16^k}\label{1644H954k443k431}\tag{67}\] and \(S_4\) stands for \[\sum_{k=1}^{\infty}\frac{\binom{4k}{k}H_{4k}(-176k^3-48k^2+80k+24)}{(3k+1)(3k+2)16^k}+\sum_{k=0}^{\infty}\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}(H_{4k+4}-H_{4k})}{(3k+1)(3k+2)16^k}.\label{1644H954k443k431443k432}\tag{68}\]

In light of 65 and the facts that \(S_3=0=S_4\), we have \[\begin{align} &\;\;\;\frac{1}{5}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{4k}-\frac{449k-275}{2}-\frac{85}{12k}\right) +\sum_{k=1}^\infty\frac{\binom{4k}{k}\big((11k^2+8k+1)H_{4k}+\frac{79}{4}k+\frac{65}{4}+\frac{17}{6k}\big)}{(3k+1)(3k+2)16^k}\\ &=\frac{1}{5}\sum_{k=1}^\infty\frac{\binom{4k}{k}(-\frac{449k-275}{2}-\frac{85}{12k})}{16^k} +\frac{3}{40}\left(\sum_{k=1}^\infty\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}}{(3k+1)16^k}(H_{4k+4}-H_{4k})+50\right)\\ &\;\;\;+\sum_{k=1}^\infty\frac{\binom{4k}{k}\left(\frac{79}{4}k+\frac{65}{4}+\frac{17}{6k}\right)}{(3k+1)(3k+2)16^k}-\frac{3}{8}\left(\sum_{k=1}^\infty\frac{8(2k+1)(4k+1)(4k+3)\binom{4k}{k}}{(3k+1)(3k+2)16^k}(H_{4k+4}-H_{4k})+25\right)\\ &=-\frac{9}{20}\sum_{k=1}^\infty\frac{\binom{4k}{k}(770k^4+1134k^3+82k^2-381k-105)}{(1+k)(1+3k)(2+3k)16^k}-\frac{45}{8}\\ &=-\frac{9}{20}\times\frac{117}{2}-\frac{45}{8}=-\frac{639}{20} \end{align}\] with the aid of 63 . Thus, we have \[\begin{align} &\;\sum_{k=1}^\infty \frac{\binom{4k}{k}((11k^2+8k+1)H_{4k}+\frac{79}{4}k+\frac{65}{4}+\frac{17}{6k})}{(3k+1)(3k+2)16^k} \\=&\;-\frac{1}{5}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{4k}-\frac{449k-275}{2}-\frac{85}{12k}\right)-\frac{639}{20}. \end{align}\] So 66 follows from 8 .0◻

. Putting \(m=16\) and \(\psi(k)=H_k\) in 57 and 58 and letting \(n\to +\infty\), we then obtain \[\sum_{k=1}^{\infty}\frac{\binom{4k}{k}\big(H_k(-176k^3+384k^2+224k+24)+48(3k-1)(3k+1)\big)}{(3k+1)16^k} =0\label{1644H95k443k431}\tag{69}\] and \[\sum_{k=1}^{\infty}\frac{\binom{4k}{k}\big(H_k(-176k^3-48k^2+80k+24)+48(3k+1)(3k+2)\big)}{(3k+1)(3k+2)16^k}=0.\label{1644H95k443k431443k432}\tag{70}\]

In light of 65 , 69 and 70 , we have \[\begin{align} &\;\;\;\frac{1}{5}\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(P(k)H_{k}-54k+108-\frac{10}{3k}\right)+\sum_{k=1}^\infty\frac{\binom{4k}{k}\left((11k^2+8k+1)H_k+6k+6+\frac{4}{3k}\right)}{(3k+1)(3k+2)16^k}\\ &=\frac{1}{5}\sum_{k=1}^\infty\frac{\binom{4k}{k}(-54k+108-\frac{10}{3k})}{16^k} +\frac{3}{40}\sum_{k=1}^\infty\frac{48(3k-1)\binom{4k}{k}}{16^k}-\frac{3}{8}\sum_{k=1}^\infty\frac{48\binom{4k}{k}}{16^k}\\ &\;\;\;+\sum_{k=1}^\infty\frac{\binom{4k}k(6k+6+\frac{4}{3k})}{(3k+1)(3k+2)16^k}\\ &=\sum_{k=1}^\infty \frac{\binom{4k}{k}}{16^k}\left(\frac{1}{5}(-54k+108-\frac{10}{3k})+\frac{18}{5}(3k-1)-18-\frac{2}{k}\right)=0. \end{align}\] So 9 follows from 5 .

Similarly, in light of 9 and 64 , we have \[\begin{align} &\;\;\;\sum_{k=0}^\infty\frac{\binom{4k}{k}((11k^2+8k+1)(H_{2k}-\frac{5}{4}H_k)+4k+1)}{(3k+1)(3k+2)16^k}-\frac{1}{2}\\ &=\sum_{k=1}^\infty\frac{\binom{4k}{k}\big((11k^2+8k+1)H_{2k}+\frac{23}{2}k+\frac{17}{2}+\frac{5}{3k}\big)}{(3k+1)(3k+2)16^k} -\sum_{k=1}^\infty\frac{\binom{4k}{k}\frac{5}{4}((11k^2+8k+1)H_k+6k+6+\frac{4}{3k})}{(3k+1)(3k+2)16^k}\\ &=\left(\frac{8}{3}\log2-\frac{1}{2}\right)-\frac{5}{4}\times \frac{4}{3}\log2=\log2-\frac{1}{2} \end{align}\] Similarly, in light of 64 and 66 , we have \[\begin{align} &\;\;\;\sum_{k=0}^\infty\frac{\binom{4k}{k}\big((11k^2+8k+1)(10H_{4k}-17H_{2k})+2k+18\big)}{(3k+1)(3k+2)16^k}\\ &=10\sum_{k=1}^\infty\frac{\binom{4k}{k}\big((11k^2+8k+1)H_{4k}+\frac{79}{4}k+\frac{65}{4}+\frac{17}{6k}\big)}{(3k+1)(3k+2)16^k} \\ &\;\;\;-17\sum_{k=1}^\infty\frac{\binom{4k}{k}\big((11k^2+8k+1)H_{2k}+\frac{23}{2}k+\frac{17}{2}+\frac{5}{3k}\big)}{(3k+1)(3k+2)16^k}+9 \\ &=10\times\left(\frac{16}{3}\log2-\frac{7}{4}\right)-17\times \left(\frac{8}{3}\log2-\frac{1}{2}\right)+9=8\log2. \end{align}\] So 10 follows from 5 , and 64 . 11 follows from 64 and 66 .

In view of the above, we have completed the proof of Theorem 2. 0◻

5 Proof of Theorem \(\ref{3}\)↩︎

. We just prove 71 in details. Other formulas in the lemma can be proved similarly.

Let \(\beta=f(-1/{256})\). Applying 27 with \(x=-1/256\), we find that \[0=7(28\beta^4-18\beta^2-8\beta-1)=(14\beta^2-7\sqrt{2}\beta-1-2\sqrt{2})(14\beta^2+7\sqrt{2} \beta-1+2\sqrt{2}).\] As \(\beta\in \mathbb{R}\), we have \(14\beta^2+7\sqrt{2} \beta-1+2\sqrt{2}\not=0\) and hence \[14\beta^2-7\sqrt{2}\beta-1-2\sqrt{2}=0. \label{-25644eq}\tag{76}\] Thus, with the aids of 29 , 32 , 60 and 76 , we deduce that \[\begin{align} &\;\;\;\frac{64}{5}\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(182k+5)}{(-256)^k}-\frac{36}{5}\sum_{k=0}^{\infty}\frac{\binom{4k}{k}}{(3k+1)(-256)^k} -\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(8k+2)}{(3k+1)(3k+2)(-256)^k}\\ &=\frac{64}{5}\left(-\frac{182}{256}\cdot\frac{64\beta^5}{(3\beta+1)^2}+5\beta\right)-\frac{36}{5}\frac{4\beta}{3\beta+1}-\left(\frac{4\beta}{3\beta+1}\right)^2-\frac{72}{5}\sqrt{2} +\frac{72}{5}\sqrt{2}\\ &=-\frac{8(14\beta^2-7\sqrt{2}\beta-1-2\sqrt{2})}{35(3\beta+1)^2}\lambda+\frac{72}{5}\sqrt{2} =\frac{72}{5}\sqrt{2}, \end{align}\] where \[\lambda=182\beta^3+ 91 \sqrt{2}\beta^2+ (-76 + 26 \sqrt{2})\beta-36 + 9 \sqrt{2}.\] This proves the desired 71 . 0◻

. Putting \(m=-256\) and \(\psi(k)=H(k)\) in 57 and 58 and letting \(n\to +\infty\), we then obtain \[\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(H(k)(896k^3+48k^2-74k+3)+2)}{(3k+1)(-256)^k}=0 \label{-256441}\tag{77}\] and \[\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(H(k)(896k^3+912k^2+214k+3)+2)}{(3k+1)(3k+2)(-256)^k}=0. \label{-256442}\tag{78}\] Observe that \[\begin{align} &\;\;\;\sum_{k=0}^\infty\frac{\binom{4k}{k}(112k^2+110k+23)H(k)}{(3k+1)(3k+2)(-256)^k}+\frac{64}{5}\sum_{k=0}^\infty\frac{\binom{4k}{k}(224k^2-86k+1)H(k)}{(-256)^k}\\ &=\frac{48}{5}\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(896k^3+48k^2-74k+3)H(k)}{(3k+1)(-256)^k} -3\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(896k^3+912k^2+214k+3)H(k)}{(3k+1)(3k+2)(-256)^k}. \end{align}\] Combining this with 77 and 78 , we obtain \[\begin{align} &\;\;\;\sum_{k=0}^\infty\frac{\binom{4k}{k}((112k^2+110k+23)H(k)+28k+16)}{(3k+1)(3k+2)(-256)^k}+ \frac{64}{5}\sum_{k=0}^\infty\frac{\binom{4k}{k}(224k^2-86k+1)H(k)}{(-256)^k}\\ &=\sum_{k=0}^\infty \frac{\binom{4k}{k}(28k+16)}{(3k+1)(3k+2)(-256)^k}-\frac{96}{5}\sum_{k=0}^{\infty}\frac{\binom{4k}{k}}{(3k+1)(-256)^k}+6\sum_{k=0}^{\infty}\frac{\binom{4k}{k}}{(3k+1)(3k+2)(-256)^k}\\ &=-\frac{36}{5}\sum_{k=0}^{\infty}\frac{\binom{4k}{k}}{(3k+1)(-256)^k}-\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(8k+2)}{(3k+1)(3k+2)(-256)^k}\\ &=\frac{72}{5}\sqrt{2}-\frac{64}{5}\sum_{k=0}^{\infty}\frac{\binom{4k}{k}(182k+5)}{(-256)^k}. \end{align}\] with the aid of 71 . Thus \[\begin{align} &\;\;\;\sum_{k=0}^\infty\frac{\binom{4k}{k}((112k^2+110k+23)H(k)+28k+16)}{(3k+1)(3k+2)(-256)^k}\\ &=\frac{72}{5}\sqrt{2}-\frac{64}{5}\sum_{k=0}^\infty\frac{\binom{4k}{k}((224k^2-86k+1)H(k)+182k+5)}{(-256)^k}.\label{-25644lemma144h} \end{align}\tag{79}\] and hence 13 is equivalent to 12 . Similarly, 14 and 15 are equivalent, and 16 and 17 are equivalent. Also, 18 and 19 are equivalent, and 20 and 21 are equivalent.

Now it remains to prove 12 , 14 , 16 , 18 and 20 . As this can be done in the way we prove Theorem 3 , we omit the details. 0◻

. There are no competing interests to declare.

References↩︎