Some combinatorial identities containing central binomial coefficients or Catalan numbers*

Abstract

In the article, by virtue of Maclaurin's expansions of the arcsine function and its square and cubic, the authors

1. give a short proof of a sum formula of a Maclaurin's series with coefficients containing reciprocals of the Catalan numbers;

2. establish four sum formulas for finite sums containing the ratio or product of two central binomial coefficients or the Catalan numbers.

The instant proof simplifies discussions in the journal papers: College Math. J. 43 (2012), no. 2, 141–146; Amer. Math. Monthly 121 (2014), no. 3, 267–267; Amer. Math. Monthly 123 (2016), no. 4, 405–406; Elem. Math. 71 (2016), no. 3, 109–121; and Mathematics 5 (2017), no. 3, Article 40, 31 pages.

Keywords:

• Catalan number
• arcsine
• square
• cubic
• ratio
• product
• sum formula
• central binomial coefficient
• finite sum
• Maclaurin's expansion

Mathematics Subject Classifications:

• Primary 05A10
• 11B65
• Secondary 11B83
• 26A09
• 33B10
• 41A58

1. Maclaurin's expansions of the powers of the arcsine function

The sequence of central binomial coefficients $(\genfrac{}{}{0ex}{}{2\ell }{\ell })$ for $\ell \ge 0$ is classical, simple, and elementary. This sequence has attracted many mathematicians who have published a number of papers such as [1–11].

Maclaurin's expansion of $\arcsin v$ can be written as (1) $\arcsin v=\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(1) See [12, 4.4.40] and [13, p. 121, 6.41.1].

Maclaurin's expansion of $(\arcsin v{)}^2$ can be formulated as (2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) See [13, p. 122, 6.42.1], [14, pp. 262–263, Proposition 15], [15, pp. 50–51 and p. 287], [16, p. 384], [17, Lemma 2], [18, p. 308], [19, pp. 88–90], [20, p. 61, 1.645], [21, p. 453], [22, Section 6.3], [23, p. 59, (2.56)], or [24, p. 676, (2.2)].

Maclaurin's expansion of $(\arcsin v{)}^3$ can be arranged as (3) $(\arcsin v{)}^3=3!\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{v^{2\ell +3}}{2\ell +3},|v|\le 1,$(3) where $C_{\ell }=\frac{1}{\ell +1}(\genfrac{}{}{0ex}{}{2\ell }{\ell })$for $\ell \ge 0$ denotes the Catalan numbers [25–28]. See [13, p. 122, 6.42.2], [14, pp. 262–263, Proposition 15], [29, p. 188, Example 1], [18, p. 308], [19, pp. 88–90], or [20, p. 61, 1.645].

Differentiating (1(1) $\arcsin v=\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(1) ), (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ), and (3(3) $(\arcsin v{)}^3=3!\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{v^{2\ell +3}}{2\ell +3},|v|\le 1,$(3) ) and simplifying result in (4) $\begin{array}{rl}\frac{1}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell },\end{array}$(4) (5) $\begin{array}{rl}\frac{\arcsin v}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1},\end{array}$(5) (6) $\begin{array}{rl}\frac{(\arcsin v{)}^2}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}.\end{array}$(6) Maclaurin's expansion (5(5) $\begin{array}{rl}\frac{\arcsin v}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1},\end{array}$(5) ) was also listed or recovered in [13, p. 122, 6.42.5], [16, p. 384], [30, p. 161], [21, p. 452, Theorem], and [22, Section 6.3, Theorem 21, Sections 8 and 9].

Maclaurin's expansions (1(1) $\arcsin v=\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(1) ), (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ), (3(3) $(\arcsin v{)}^3=3!\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{v^{2\ell +3}}{2\ell +3},|v|\le 1,$(3) ), (4(4) $\begin{array}{rl}\frac{1}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell },\end{array}$(4) ), (5(5) $\begin{array}{rl}\frac{\arcsin v}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1},\end{array}$(5) ), and (6(6) $\begin{array}{rl}\frac{(\arcsin v{)}^2}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}.\end{array}$(6) ) have been employed in [17, 31] to establish sum formulas for finite sums containing the quantities $(\genfrac{}{}{0ex}{}{2m}{m})(\genfrac{}{}{0ex}{}{2(\ell -m)}{\ell -m})\mathrm{o}\mathrm{r}\frac{(\genfrac{}{}{0ex}{}{2m}{m})}{(\genfrac{}{}{0ex}{}{2(\ell -m+1)}{\ell -m+1})}$for $\ell \ge m\ge 0$.

In 1922, Maclaurin's expansion of $\sqrt{1-v^2}\arcsin v$ with minor errors was listed in [13, p. 122, 6.42.4] and we now correct it as (7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) Recently, Maclaurin's series expansions of the functions $(\arcsin t{)}^m,{\left(\frac{\arcsin t}{t}\right)}^m,\frac{(\arcsin t{)}^m}{\sqrt{1-t^2}}$ for $m\in \mathbb{N}$ have been reviewed and surveyed in [32]. Hereafter, very nice Maclaurin's and Taylors's series expansions of the functions $\begin{array}{rl}& {\left(\frac{\arcsin t}{t}\right)}^{\alpha },\frac{(\arcsin t{)}^{\alpha }}{\sqrt{1-t^2}},{\left(\frac{\mathrm{arcsinh}t}{t}\right)}^m,\\ & {\left[\frac{(\arccos x{)}^2}{2(1-x)}\right]}^{\alpha },{\left[\frac{(\mathrm{arccosh}x{)}^2}{2(1-x)}\right]}^m,{\left(\frac{2\arccos t}{\pi }\right)}^{\alpha }\end{array}$were discovered in the papers [33–36], where $m\in \mathbb{N}$ and $\alpha \in \mathbb{R}$.

In this paper, by virtue of the above seven Maclaurin's expansions (1(1) $\arcsin v=\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(1) ), (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ), (3(3) $(\arcsin v{)}^3=3!\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{v^{2\ell +3}}{2\ell +3},|v|\le 1,$(3) ), (4(4) $\begin{array}{rl}\frac{1}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell },\end{array}$(4) ), (5(5) $\begin{array}{rl}\frac{\arcsin v}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1},\end{array}$(5) ), (6(6) $\begin{array}{rl}\frac{(\arcsin v{)}^2}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}.\end{array}$(6) ), and (7(7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) ), we will

1. give a short proof of a sum formula of the series $\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{v^{\ell }}{C_{\ell }}$for $0\le v\le 4$;

2. establish four sum formulas for finite sums containing the ratio or product of two central binomial coefficients or the Catalan numbers $C_{\ell }$.

For details on our main results, please read Sections 2 and 3 below.

2. A short proof of a sum formula

The sum formulas (8(8) $\sum _{\ell =0}^{\mathrm{\infty }}\frac{v^{\ell }}{C_{\ell }}=\frac{2(8+v)}{(4-v{)}^2}+\frac{24\sqrt{v}}{(4-v{)}^{5/2}}\arcsin \frac{\sqrt{v}}{2}$(8) ) and (9(9) $\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{v^{\ell }}{C_{\ell }}=\frac{2(8-v)}{(4+v{)}^2}+\frac{24\sqrt{v}}{(4+v{)}^{5/2}}\ln \frac{\sqrt{4+v}-\sqrt{v}}{2}.$(9) ) in Theorem 2.1 below have been proved in [37–41] and [22, pp. 17–28, Sections 6–9] by spending much space on complicated and technical arguments.

Theorem 2.1

[37–41] and [22, pp. 17–28, Sections 6–9]

For $0\le v\le 4$, (8) $\sum _{\ell =0}^{\mathrm{\infty }}\frac{v^{\ell }}{C_{\ell }}=\frac{2(8+v)}{(4-v{)}^2}+\frac{24\sqrt{v}}{(4-v{)}^{5/2}}\arcsin \frac{\sqrt{v}}{2}$(8) and (9) $\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{v^{\ell }}{C_{\ell }}=\frac{2(8-v)}{(4+v{)}^2}+\frac{24\sqrt{v}}{(4+v{)}^{5/2}}\ln \frac{\sqrt{4+v}-\sqrt{v}}{2}.$(9)

By virtue of Maclaurin's expansion (7(7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) ), we now give a marvelous, short, instant, simple, and elementary proof of the sum formulas (8(8) $\sum _{\ell =0}^{\mathrm{\infty }}\frac{v^{\ell }}{C_{\ell }}=\frac{2(8+v)}{(4-v{)}^2}+\frac{24\sqrt{v}}{(4-v{)}^{5/2}}\arcsin \frac{\sqrt{v}}{2}$(8) ) and (9(9) $\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{v^{\ell }}{C_{\ell }}=\frac{2(8-v)}{(4+v{)}^2}+\frac{24\sqrt{v}}{(4+v{)}^{5/2}}\ln \frac{\sqrt{4+v}-\sqrt{v}}{2}.$(9) ).

Proof of Theorem 2.1.

We can write Maclaurin's expansion (7(7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) ) as (10) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(10) Maclaurin's expansion (10(10) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(10) ) can be further rewritten in terms of the Catalan numbers $C_{\ell }$ as (11) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)\ell C_{\ell -1}}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(11) Differentiating three times on both sides of (11(11) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)\ell C_{\ell -1}}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(11) ) and simplifying yield (12) $\frac{3v\arcsin v}{(1-v^2{)}^{5/2}}+\frac{v^2+2}{(1-v^2{)}^2}=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{C_{\ell }}{v}^{2\ell },|v|\le 1.$(12) Taking $v^2=\frac{u}{4}$ for $0\le u\le 4$ in (12(12) $\frac{3v\arcsin v}{(1-v^2{)}^{5/2}}+\frac{v^2+2}{(1-v^2{)}^2}=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{C_{\ell }}{v}^{2\ell },|v|\le 1.$(12) ) and simplifying give the sum formula (8(8) $\sum _{\ell =0}^{\mathrm{\infty }}\frac{v^{\ell }}{C_{\ell }}=\frac{2(8+v)}{(4-v{)}^2}+\frac{24\sqrt{v}}{(4-v{)}^{5/2}}\arcsin \frac{\sqrt{v}}{2}$(8) ).

Letting $v^2=-\frac{u}{4}$ for $0\le u\le 4$ in (12(12) $\frac{3v\arcsin v}{(1-v^2{)}^{5/2}}+\frac{v^2+2}{(1-v^2{)}^2}=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{C_{\ell }}{v}^{2\ell },|v|\le 1.$(12) ), we find (13) $\frac{3}{(1+\frac{u}{4}{)}^{5/2}}\frac{j\sqrt{u}}{2}\arcsin \frac{j\sqrt{u}}{2}+\frac{2-\frac{u}{4}}{(1+\frac{u}{4}{)}^2}=2\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{u^{\ell }}{C_{\ell }},$(13) where $j=\sqrt{-1}$ is the imaginary unit in complex analysis. Making use of the logarithmic representation $\arcsin v=-j\ln (\sqrt{1-v^2}+iv),v^2\le 1$in [12, p. 80, 4.4.26] and [42, p. 119, 4.23.19], we obtain $\arcsin \frac{j\sqrt{u}}{2}=-j\ln [\sqrt{1-(\frac{j\sqrt{u}}{2}{)}^2}+j\frac{j\sqrt{u}}{2}]=-j\ln \frac{\sqrt{4+u}-\sqrt{u}}{2}.$Substituting this equality into (13(13) $\frac{3}{(1+\frac{u}{4}{)}^{5/2}}\frac{j\sqrt{u}}{2}\arcsin \frac{j\sqrt{u}}{2}+\frac{2-\frac{u}{4}}{(1+\frac{u}{4}{)}^2}=2\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{u^{\ell }}{C_{\ell }},$(13) ) and simplifying produce $\frac{3}{(1+\frac{u}{4}{)}^{5/2}}\frac{\sqrt{u}}{2}\ln \frac{\sqrt{4+u}-\sqrt{u}}{2}+\frac{2-\frac{u}{4}}{(1+\frac{u}{4}{)}^2}=2\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{u^{\ell }}{C_{\ell }}$which is equivalent to $2\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{u^{\ell }}{C_{\ell }}=\frac{4(8-u)}{(4+u{)}^2}+\frac{48\sqrt{u}}{(4+u{)}^{5/2}}\ln \frac{\sqrt{4+u}-\sqrt{u}}{2}.$The sum formula (9(9) $\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{v^{\ell }}{C_{\ell }}=\frac{2(8-v)}{(4+v{)}^2}+\frac{24\sqrt{v}}{(4+v{)}^{5/2}}\ln \frac{\sqrt{4+v}-\sqrt{v}}{2}.$(9) ) is proved.

Remark 2.1

The short proof of Theorem 2.1 shows that the method used for proving the sum formula (9(9) $\sum _{\ell =0}^{\mathrm{\infty }}(-1{)}^{\ell }\frac{v^{\ell }}{C_{\ell }}=\frac{2(8-v)}{(4+v{)}^2}+\frac{24\sqrt{v}}{(4+v{)}^{5/2}}\ln \frac{\sqrt{4+v}-\sqrt{v}}{2}.$(9) ) is better than previous ones. Therefore, we will apply this method to establish several more formulas for finite sums containing the quantities $(\genfrac{}{}{0ex}{}{2\ell }{\ell })$ or $C_{\ell }$ in next section.

3. Four identities containing central binomial coefficients

In this section, by virtue of Maclaurin's expansions (1(1) $\arcsin v=\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(1) ), (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ), (3(3) $(\arcsin v{)}^3=3!\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{v^{2\ell +3}}{2\ell +3},|v|\le 1,$(3) ), (4(4) $\begin{array}{rl}\frac{1}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell },\end{array}$(4) ), (5(5) $\begin{array}{rl}\frac{\arcsin v}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1},\end{array}$(5) ), (6(6) $\begin{array}{rl}\frac{(\arcsin v{)}^2}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}.\end{array}$(6) ), (7(7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) ), we establish four sum formulas for finite sums containing the quantities $(\genfrac{}{}{0ex}{}{2\ell }{\ell })$ or $C_{\ell }$.

Theorem 3.1

For $\ell \in \mathbb{N}$, (14) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{2^{4q}}{(2q+1)(2q+3)}\frac{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}{(\genfrac{}{}{0ex}{}{2q}{q})}=\frac{\ell }{2(2\ell +1)}(\genfrac{}{}{0ex}{}{2\ell }{\ell }),\end{array}$(14) (15) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{1}{(2q+1)(2q+3)(\ell -q)}\frac{1}{(\genfrac{}{}{0ex}{}{2q}{q})(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}=\frac{2\ell }{(\ell +1)(2\ell +1)(\genfrac{}{}{0ex}{}{2\ell }{\ell })},\end{array}$(15) (16) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{3q+4}{(q+1)(q+2)}\frac{\left(2q\right)!!}{(2q+3)!!}=1-\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{(\ell +1{)}^2},\end{array}$(16) and (17) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{2q+1}{2^{4q}\left[2\right(\ell -q)-1]\left[2\right(\ell -q)+1]}\frac{C_q}{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\\ & =\frac{\ell (2\ell +1)C_{\ell }}{2^{4\ell -3}(2\ell +3)}\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}.\end{array}$(17)

Proof of Theorem 3.1.

Utilizing the Cauchy product of the product of two Maclaurin's expansions, we acquire $\begin{array}{rl}\arcsin v& =\frac{1}{\sqrt{1-v^2}}(\sqrt{1-v^2}\arcsin v)\\ & =\left[\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell }\right][v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1}]\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell +1}-\left[\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell }\right]\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell }}{(2\ell +1)(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{v^{2\ell +3}}{2\ell +3}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell +1}-\sum _{\ell =0}^{\mathrm{\infty }}\left[\sum _{q=0}^{\ell }\frac{2^{2(2q-\ell )}}{(2q+1)(2q+3)}\frac{(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}{(\genfrac{}{}{0ex}{}{2q}{q})}\right]v^{2\ell +3}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell +1}-\sum _{\ell =1}^{\mathrm{\infty }}\left[\sum _{q=0}^{\ell -1}\frac{2^{2(2q-\ell +1)}}{(2q+1)(2q+3)}\frac{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}{(\genfrac{}{}{0ex}{}{2q}{q})}\right]v^{2\ell +1}\\ & =v+\sum _{\ell =1}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}[(\genfrac{}{}{0ex}{}{2\ell }{\ell })-\sum _{q=0}^{\ell -1}\frac{2^{2(2q+1)}}{(2q+1)(2q+3)}\frac{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}{(\genfrac{}{}{0ex}{}{2q}{q})}]v^{2\ell +1},\end{array}$where we used (4(4) $\begin{array}{rl}\frac{1}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })v^{2\ell },\end{array}$(4) ) and (10(10) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(10) ). Comparing this with (1(1) $\arcsin v=\sum _{\ell =0}^{\mathrm{\infty }}\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(1) ) and equating the coefficient of the term $v^{2\ell +1}$ for $\ell \in \mathbb{N}$ lead to $\frac{1}{2^{2\ell }}(\genfrac{}{}{0ex}{}{2\ell }{\ell })\frac{1}{2\ell +1}=\frac{1}{2^{2\ell }}[(\genfrac{}{}{0ex}{}{2\ell }{\ell })-\sum _{q=0}^{\ell -1}\frac{2^{2(2q+1)}}{(2q+1)(2q+3)}\frac{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}{(\genfrac{}{}{0ex}{}{2q}{q})}]$which is equivalent to (14(14) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{2^{4q}}{(2q+1)(2q+3)}\frac{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}{(\genfrac{}{}{0ex}{}{2q}{q})}=\frac{\ell }{2(2\ell +1)}(\genfrac{}{}{0ex}{}{2\ell }{\ell }),\end{array}$(14) ).

Similarly, we can write $\begin{array}{rl}(\arcsin v{)}^2& =\frac{\arcsin v}{\sqrt{1-v^2}}(\sqrt{1-v^2}\arcsin v)\\ & =\left[\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1}\right][v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1}]\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +2}}{\ell +1}-\left[\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1}\right]\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell }}{(2\ell +1)(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{v^{2\ell +3}}{2\ell +3}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +2}}{\ell +1}\\ & -\sum _{\ell =0}^{\mathrm{\infty }}\left[\sum _{q=0}^{\ell }\frac{2^{2(\ell -q)+1}}{(\genfrac{}{}{0ex}{}{2(\ell -q+1)}{\ell -q+1})}\frac{1}{\ell -q+1}\frac{2^{2q}}{(2q+1)(\genfrac{}{}{0ex}{}{2q}{q})}\frac{1}{2q+3}\right]v^{2\ell +4}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +2}}{\ell +1}\\ & -\sum _{\ell =1}^{\mathrm{\infty }}{2}^{2\ell -1}\left[\sum _{q=0}^{\ell -1}\frac{1}{(\genfrac{}{}{0ex}{}{2q}{q})(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}\frac{1}{(2q+1)(2q+3)(\ell -q)}\right]v^{2\ell +2}\\ & =v^2+\sum _{\ell =1}^{\mathrm{\infty }}{2}^{2\ell -1}[\frac{4}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{\ell +1}\\ & -\sum _{q=0}^{\ell -1}\frac{1}{(\genfrac{}{}{0ex}{}{2q}{q})(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}\frac{1}{(2q+1)(2q+3)(\ell -q)}]v^{2\ell +2},\end{array}$where we used (5(5) $\begin{array}{rl}\frac{\arcsin v}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2\ell +1}}{\ell +1},\end{array}$(5) ) and (10(10) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(10) ). Comparing this with Maclaurin's expansion (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ) and equating result in $\begin{array}{r}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{(\ell +1{)}^2}=2^{2\ell -1}[\frac{4}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{\ell +1}-\sum _{q=0}^{\ell -1}\frac{1}{(\genfrac{}{}{0ex}{}{2q}{q})(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}\frac{1}{(2q+1)(2q+3)(\ell -q)}],\end{array}$which can be rearranged as (15(15) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{1}{(2q+1)(2q+3)(\ell -q)}\frac{1}{(\genfrac{}{}{0ex}{}{2q}{q})(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}=\frac{2\ell }{(\ell +1)(2\ell +1)(\genfrac{}{}{0ex}{}{2\ell }{\ell })},\end{array}$(15) ).

Direct computation gives $\begin{array}{rl}& \frac{\text{d}(\sqrt{1-v^2}\arcsin v{)}^2}{\text{d}v}=2\sqrt{1-v^2}\arcsin v-2v(\arcsin v{)}^2\\ & =2[v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1}]-2v\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2}\\ & =2v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -1}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1}-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell }}{(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{v^{2\ell +1}}{{\ell }^2}\\ & =2v-\sum _{\ell =1}^{\mathrm{\infty }}{2}^{2\ell -1}(\frac{\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{1}{2\ell +1}+\frac{2}{(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{1}{{\ell }^2})v^{2\ell +1}\\ & =2v-\sum _{\ell =1}^{\mathrm{\infty }}{2}^{2\ell -1}[\frac{1}{(4{\ell }^2-1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}+\frac{2}{{\ell }^2(\genfrac{}{}{0ex}{}{2\ell }{\ell })}]v^{2\ell +1},\end{array}$where Maclaurin's expansions (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ) and (7(7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) ) were employed. Accordingly, we obtain (18) $(\sqrt{1-v^2}\arcsin v{)}^2=v^2-\sum _{\ell =1}^{\mathrm{\infty }}\frac{3\ell +1}{\ell (\ell +1)}\frac{(2\ell -2)!!}{(2\ell +1)!!}{v}^{2\ell +2}.$(18) Maclaurin's expansion (18(18) $(\sqrt{1-v^2}\arcsin v{)}^2=v^2-\sum _{\ell =1}^{\mathrm{\infty }}\frac{3\ell +1}{\ell (\ell +1)}\frac{(2\ell -2)!!}{(2\ell +1)!!}{v}^{2\ell +2}.$(18) ) can also be derived from (7(7) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -2}\left[\right(\ell -1)!{]}^2}{(2\ell -1)!}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(7) ) as follows: $\begin{array}{rl}(\sqrt{1-v^2}\arcsin v{)}^2& =(1-v^2)(\arcsin v{)}^2\\ & =(1-v^2)\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2}-\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +2)}}{(\ell +1{)}^2}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2}-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2\ell -1}}{(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{v^{2(\ell +1)}}{{\ell }^2}\\ & =v^2+\sum _{\ell =1}^{\mathrm{\infty }}[\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{(\ell +1{)}^2}-\frac{2^{2\ell -1}}{(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{1}{{\ell }^2}]v^{2(\ell +1)}\\ & =v^2-\sum _{\ell =1}^{\mathrm{\infty }}\frac{3\ell +1}{\ell (\ell +1)}\frac{(2\ell -2)!!}{(2\ell +1)!!}{v}^{2\ell +2},\end{array}$where we used Maclaurin's expansion (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ).

Employing Maclaurin's expansion (18(18) $(\sqrt{1-v^2}\arcsin v{)}^2=v^2-\sum _{\ell =1}^{\mathrm{\infty }}\frac{3\ell +1}{\ell (\ell +1)}\frac{(2\ell -2)!!}{(2\ell +1)!!}{v}^{2\ell +2}.$(18) ) yileds $\begin{array}{rl}(\arcsin v{)}^2& =\frac{1}{1-v^2}(\sqrt{1-v^2}\arcsin v{)}^2\\ & =\left(\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell }\right)[v^2-\sum _{\ell =1}^{\mathrm{\infty }}\frac{3\ell +1}{\ell (\ell +1)}\frac{(2\ell -2)!!}{(2\ell +1)!!}{v}^{2\ell +2}]\\ & =\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell +2}-\left(\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell }\right)\sum _{\ell =1}^{\mathrm{\infty }}\frac{3\ell +1}{\ell (\ell +1)}\frac{(2\ell -2)!!}{(2\ell +1)!!}{v}^{2\ell +2}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell +2}-\left(\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell }\right)\sum _{\ell =0}^{\mathrm{\infty }}\frac{3\ell +4}{(\ell +1)(\ell +2)}\frac{\left(2\ell \right)!!}{(2\ell +3)!!}{v}^{2\ell +4}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell +2}-\sum _{\ell =0}^{\mathrm{\infty }}\left[\sum _{q=0}^{\ell }\frac{3q+4}{(q+1)(q+2)}\frac{\left(2q\right)!!}{(2q+3)!!}\right]v^{2\ell +4}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}{v}^{2\ell +2}-\sum _{\ell =1}^{\mathrm{\infty }}\left[\sum _{q=0}^{\ell -1}\frac{3q+4}{(q+1)(q+2)}\frac{\left(2q\right)!!}{(2q+3)!!}\right]v^{2\ell +2}\\ & =v^2+\sum _{\ell =1}^{\mathrm{\infty }}[1-\sum _{q=0}^{\ell -1}\frac{3q+4}{(q+1)(q+2)}\frac{\left(2q\right)!!}{(2q+3)!!}]v^{2\ell +2}.\end{array}$Comparing this with (2(2) $(\arcsin v{)}^2=\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{v^{2(\ell +1)}}{(\ell +1{)}^2},|v|\le 1.$(2) ) and equating conclude $\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{(\ell +1{)}^2}=1-\sum _{q=0}^{\ell -1}\frac{3q+4}{(q+1)(q+2)}\frac{\left(2q\right)!!}{(2q+3)!!}.$The sum formula (16(16) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{3q+4}{(q+1)(q+2)}\frac{\left(2q\right)!!}{(2q+3)!!}=1-\frac{2^{2\ell +1}}{(\genfrac{}{}{0ex}{}{2(\ell +1)}{\ell +1})}\frac{1}{(\ell +1{)}^2},\end{array}$(16) ) is thus proved.

Finally, we write $\begin{array}{rl}(\arcsin v{)}^3& =\frac{(\arcsin v{)}^2}{\sqrt{1-v^2}}(\sqrt{1-v^2}\arcsin v)\\ & =\left(\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}\right)\\ & \times [v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1}]\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +3}\\ & -\left(\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}\right)\sum _{\ell =0}^{\mathrm{\infty }}\frac{2^{2\ell }}{(2\ell +1)(\genfrac{}{}{0ex}{}{2\ell }{\ell })}\frac{v^{2\ell +3}}{2\ell +3}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +3}\\ & -\sum _{\ell =0}^{\mathrm{\infty }}(\sum _{q=0}^{\ell }\frac{2q+1}{2^{2q}}{C}_q\left[\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\right]\\ & \times \frac{2^{2(\ell -q)}}{\left[2\right(\ell -q)+1](\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})}\frac{1}{2(\ell -q)+3})v^{2\ell +5}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +3}\\ & -\sum _{\ell =0}^{\mathrm{\infty }}{2}^{2\ell }(\sum _{q=0}^{\ell }\frac{2q+1}{2^{4q}\left[2\right(\ell -q)+1]\left[2\right(\ell -q)+3]}\\ & \times \left[\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\right]\frac{C_q}{(\genfrac{}{}{0ex}{}{2(\ell -q)}{\ell -q})})v^{2\ell +5}\\ & =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +3}-\sum _{\ell =1}^{\mathrm{\infty }}{2}^{2(\ell -1)}\\ & \times (\sum _{q=0}^{\ell -1}\frac{2q+1}{2^{4q}\left[2\right(\ell -q)-1]\left[2\right(\ell -q)+1]}\\ & \times \left[\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\right]\frac{C_q}{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})})v^{2\ell +3}\\ & =v^3+\sum _{\ell =1}^{\mathrm{\infty }}(\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]-2^{2(\ell -1)}\\ & \times \sum _{q=0}^{\ell -1}\frac{2q+1}{2^{4q}\left[2\right(\ell -q)-1]\left[2\right(\ell -q)+1]}\\ & \times \left[\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\right]\frac{C_q}{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})})v^{2\ell +3},\end{array}$where we used (6(6) $\begin{array}{rl}\frac{(\arcsin v{)}^2}{\sqrt{1-v^2}}& =\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]v^{2\ell +2}.\end{array}$(6) ) and (10(10) $\sqrt{1-v^2}\arcsin v=v-\sum _{\ell =1}^{\mathrm{\infty }}\frac{2^{2(\ell -1)}}{(2\ell -1)(\genfrac{}{}{0ex}{}{2(\ell -1)}{\ell -1})}\frac{v^{2\ell +1}}{2\ell +1},|v|\le 1.$(10) ). Comparing this with (3(3) $(\arcsin v{)}^3=3!\sum _{\ell =0}^{\mathrm{\infty }}\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{v^{2\ell +3}}{2\ell +3},|v|\le 1,$(3) ) and equating give $\begin{array}{rl}& 3!\frac{2\ell +1}{2^{2\ell +1}}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\frac{1}{2\ell +3}=\frac{2\ell +1}{2^{2\ell }}{C}_{\ell }\left[\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}\right]\\ & -2^{2(\ell -1)}\sum _{q=0}^{\ell -1}\frac{2q+1}{2^{4q}\left[2\right(\ell -q)-1]\left[2\right(\ell -q)+1]}\left[\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\right]\frac{C_q}{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}\end{array}$which can be rearranged as (17(17) $\begin{array}{rl}& \sum _{q=0}^{\ell -1}\frac{2q+1}{2^{4q}\left[2\right(\ell -q)-1]\left[2\right(\ell -q)+1]}\frac{C_q}{(\genfrac{}{}{0ex}{}{2(\ell -q-1)}{\ell -q-1})}\sum _{\ell =0}^q\frac{1}{(2\ell +1{)}^2}\\ & =\frac{\ell (2\ell +1)C_{\ell }}{2^{4\ell -3}(2\ell +3)}\sum _{q=0}^{\ell }\frac{1}{(2q+1{)}^2}.\end{array}$(17) ).

Acknowledgments

The authors appreciate anonymous referees for their careful reading and valuable comments on the original version of this paper.

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Funding

Dr. Dongkyu Lim, the third and corresponding author, was supported by the National Research Foundation of Korea under Grant NRF-2021R1C1C1010902, Republic of Korea.