Closed form of an infinite series

Question

Does the following infinite series have a closed form? $$ \sum_{n=1}^{\infty} {(-1)^n \frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} \sin(\frac{2\pi n}{3})} $$

Answer 1

Q: Does the following infinite series have a closed form?

It does, according to Mathematica: $$\sum_{n=1}^{\infty} {(-1)^n \frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} \sin\left(\frac{2\pi n}{3}\right)}=-2^{-10/3} \Gamma \left(\tfrac{1}{3}\right)\Big[2 \sqrt{3}+9\pi^{-1} \, _2F_1\left(\tfrac{2}{3},\tfrac{2}{3};\tfrac{5}{3};-1\right)\hspace{-1mm}\Big].$$

Answer 2

Denote $c_n:={(-1)^n \frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} \sin(\frac{2\pi n}{3})}$ the $n$-th term of the series. We have for all $k\ge0$ $$c_{3k}=0,$$ $$c_{3k+1}= (-1)^{k+1}\frac{\Gamma\big(k+\frac23\big)}{\Gamma\big(k+\frac43\big)} \frac{\sqrt 3}2= (-1)^{k+1}\frac{\sqrt 3}{2\Gamma(\frac23)}B\Big( k+\frac23, \frac23\Big) =$$$$= (-1)^{k+1}\frac{\sqrt 3}{2\Gamma(\frac23)}\int_0^1x^{k-\frac13}(1-x)^{-\frac13}dx$$ $$c_{3k+2}= (-1)^{k+1}\frac{\Gamma\big(k+1\big)}{\Gamma\big(k+\frac53\big)} \frac{\sqrt 3}2= (-1)^{k+1} \frac{\sqrt 3}{2\Gamma(\frac23)}B\Big( k+1, \frac23\Big)=$$$$= (-1)^{k+1}\frac{\sqrt 3}{2\Gamma(\frac23)}\int_0^1x^k(1-x)^{-\frac13}dx$$

We can group three consecutive terms (note incidentally that the series is not absolutely convergent, since $|c_n|\sim Cn^{-2/3}$ for $n\neq0 \text{ mod }3$). Thus $$\sum_{n=1}^\infty c_n=\frac{\sqrt 3}{2\Gamma(\frac23)}\sum_{k\ge0}(-1)^{k+1}\int_0^1x^k(1+x^{-\frac13})(1-x)^{-\frac13}dx=$$$$= -\frac{\sqrt 3}{2\Gamma(\frac23)} \int_0^1 (1+x)^{-1}(1+x^{-\frac13})(1-x)^{-\frac13}dx,$$ where the sum under the integral sign is allowed by Beppo Levi's monotone convergence theorem, now further grouping two consecutive terms, whose sum is non-negative.

Computing the integral by WolframAlpha I got

$$\sum_{n=1}^\infty c_n=\frac{-3\pi+ 3\sqrt3 \log\big(\sqrt[3]2-1\big) + 6 \arctan\Big(\frac{1+\sqrt[3]4}{\sqrt3} \Big)}{4\sqrt[3]2\Gamma\big(\frac23\big)} $$ which is $−1.544563279\dots$, according to the Maple evaluation of the above integral integral by Timothy Budd and to the computations in the previous answers.

Answer 3

Generalize to trivialize! - D. Zeilberger. In view of this, consider the following three functions \begin{align*} F_0(n,k):=\frac{z^{3k}}{\binom{3n+k-1}{2n}}, \qquad F_1(n,k):=\frac{z^{3k-1}}{\binom{3n+k-\frac43}{2n}}, \qquad F_2(n,k):=\frac{z^{3k-2}}{\binom{3n+k-\frac53}{2n}}. \end{align*} For convenience, let's modify each of them in the manner \begin{align*} \widehat{F}_0(n,k):=\frac{F_0(n,k)z^{9n}}{n(z^3-1)^{2n}}, \qquad \widehat{F}_1(n,k):=\frac{F_1(n,k)z^{9n}}{n(z^3-1)^{2n}}, \qquad \widehat{F}_2(n,k):=\frac{F_2(n,k)z^{9n}}{n(z^3-1)^{2n}}. \end{align*} Zeilberger's algorithm cranks out the corresponding companion functions in the form \begin{align*} &\widehat{G}_0(n,k):=\widehat{F}_0(n,k) \,\,\, \times\\ &\frac{(4n^2+2n)z^6 -(3n+k+1)(5n+k)z^3+(3n+k)(3n+k+1)}{(3n + k+1)(3n + k)(z^3-1)^2}, \\ &\widehat{G}_1(n,k):=\widehat{F}_1(n,k) \,\,\, \times \\ &\frac{(36n^2 + 18n)z^6 - 9(3n + k + \frac23)(5n - \frac13 + k)z^3 + 9(3n + k - \frac13)(3n + k + \frac23)}{(9n + 3k + 2)(9n + 3k - 1)(z^3 - 1)^2}, \\ &\widehat{G}_2(n,k):=\widehat{F}_2(n,k) \,\,\, \times \\ &\frac{(36n^2 + 18n)z^6 - 9(3n + k + \frac13)(5n - \frac23 + k)z^3 + 9(3n + k + \frac13)(3n + k -\frac23)}{(9n + 3k + 1)(9n + 3k - 2)(z^3 - 1)^2}. \end{align*} Now, check that $\widehat{F}_j(n+1,k)-\widehat{F}_j(n,k)=\widehat{G}_j(n,k+1)-\widehat{G}_j(n,k)$ for $j=0, 1, 2$. Next, sum both sides of the three equations over all integers $k\geq1$. Clearly, $\sum_{k\geq1}\widehat{F}_j(n+1,k)-\sum_{k\geq1}\widehat{F}_j(n,k)=-\widehat{G}_j(n,1)$. Multiply through by the factor $(z^3-1)^{2n+2}$ to write (for $j=0, 1, 2$) that \begin{align*} \sum_{k\geq1}\frac{F_j(n+1,k)z^{9n+9}}{n+1}-\sum_{k\geq1}\frac{F_j(n,k)z^{9n}(z^3-1)^2}n &=-(z^3-1)^{2n+2}\,\widehat{G}_j(n,1). \end{align*} The choice $z=-e^{\frac{2\pi i}3}$ implies $z^3=1$ and hence we can give even more detail: \begin{align*} \sum_{k\geq1}\frac1{(n+1)\binom{3n+k+2}{2n+2}}&=2\,\frac{(n+1)!\,(2n)!}{(3n+2)!}, \\ \sum_{k\geq1}\frac{-e^{-\frac{2\pi i}3}}{(n+1)\binom{3n+k+\frac53}{2n+2}}&=\frac{-e^{-\frac{2\pi i}3}6n(3n+2)}{\binom{3n-\frac13}{2n}n(9n+5)(9n+2)}, \\ \sum_{k\geq1}\frac{e^{-\frac{4\pi i}3}}{(n+1)\binom{3n+k+\frac43}{2n+2}}&=\frac{e^{-\frac{4\pi i}3}6n(3n+1)}{\binom{3n-\frac23}{2n}n(9n+4)(9n+1)}. \end{align*} Start substituting $n=-\frac23$ in each equation to obtain \begin{align*} \sum_{k\geq1}\frac{3\,\Gamma(\frac53)\,\Gamma(k+\frac13)}{\Gamma(k+1)}&=2\,\Gamma\left(\frac43\right)\Gamma\left(-\frac13\right), \\ \sum_{k\geq1}\frac{-3\,e^{-\frac{2\pi i}3}\Gamma(\frac53)\,\Gamma(k)}{\Gamma(k+\frac23)}&=6e^{-\frac{2\pi i}3}, \\ \sum_{k\geq1}\frac{3\,e^{-\frac{4\pi i}3}\Gamma(\frac53)\,\Gamma(k-\frac13)}{\Gamma(k+\frac13)}&= -\frac{3\,e^{-\frac{4\pi i}3}\Gamma\left(-\frac13\right)^2}{5\,\Gamma(-\frac53)}. \end{align*} Divide each equation by $3\Gamma(\frac53)$ and add all three equation to find $$\sum_{m\geq1}\frac{\Gamma(\frac{m}3+\frac13)}{\Gamma(\frac{m}3+1)}(-e^{\frac{2\pi i}3})^m =\frac{2\,\Gamma(\frac43)\,\Gamma(-\frac13)}{3\,\Gamma(\frac53)} +\frac{2\,e^{-\frac{2\pi i}3}}{\Gamma(\frac53)} -\frac{e^{-\frac{4\pi i}3}\Gamma(-\frac13)^2}{5\,\Gamma(\frac53)\,\Gamma(-\frac53)}.$$ Finally, if we take the imaginary part of this equation, there comes $$\sum_{m\geq1}(-1)^m\frac{\Gamma(\frac{m}3+\frac13)}{\Gamma(\frac{m}3+1)}\,\sin\left(\frac{2\pi m}3\right) =-\frac{\sqrt{3}}{\Gamma(\frac53)}-\frac{\sqrt{3}\,\Gamma(-\frac13)^2}{10\,\Gamma(\frac53)\,\Gamma(-\frac53)}.$$

I think there is a slight error but the procedure works. The main merit here could be that we don't depend on an answer in terms of hypergeometric language.

Answer 4

Here is what I can do with Maple...
Consider, instead $$ F(z) := \sum_{n=1}^{\infty} {\frac{\Gamma(\frac{1}{3}+\frac{n}{3})}{\Gamma(1+\frac{n}{3})} z^n} $$ so that the required value is $\operatorname{Im}F(-e^{i 2\pi/3})$.
Maple says
$$ F(z) = \\ {\frac {3\, \left( \Gamma \left( 2/3 \right) \right) ^{2}\sqrt {3}z}{ 2\,\pi}}+{\frac {3\,{z}^{2}}{2\,\Gamma \left( 2/3 \right) }}+{\frac {2 \,\pi\,\sqrt {3}}{3\,\Gamma \left( 2/3 \right) }{\frac {1}{\sqrt [3] {1-{z}^{3}} }}} -{\frac {2\,\pi\,\sqrt {3}}{3\,\Gamma \left( 2/3 \right) }}+{\frac {3\, \left( \Gamma \left( 2/3 \right) \right) ^{2}\sqrt {3} {z}^{4}}{4\,\pi} {\mbox{$_2$F$_1$}(1,{\frac{5}{3}};\,{\frac{7}{3}};\,{z}^{3})}}+{\frac {9\,{z}^{5}}{10\,\Gamma \left( 2/3 \right) } {\mbox{$_2$F$_1$}(1,2;\,{\frac{8}{3}};\,{z}^{3})}} $$ so that $$ \operatorname{Im}F(-e^{i2\pi/3}) = {\frac {9\, \left( \Gamma \left( 2/3 \right) \right) ^{2}}{8\,\pi} {\mbox{$_2$F$_1$}(1,{\frac{5}{3}};\,{\frac{7}{3}};\,-1)}}+{\frac {9\, \sqrt {3}}{20\,\Gamma \left( 2/3 \right) } {\mbox{$_2$F$_1$}(1,2;\,{\frac{8}{3}};\,-1)}}-{\frac {9\, \left( \Gamma \left( 2/3 \right) \right) ^{2}}{4\,\pi}}-{\frac {3\,\sqrt {3} }{4\,\Gamma \left( 2/3 \right) }} $$ Numerical evaluation $-1.54456$ agrees with Carlo.

Answer 5

$\newcommand{\Ga}{\Gamma}$This is to provide details on Will Sawin's comment.

We want to evaluate \begin{equation*} s:=\sum_{n=1}^\infty {(-1)^n \frac{\Ga(\frac13+\frac n3)}{\Ga(1+\frac n3)} \sin\frac{2\pi n}3} =\sum_{n=0}^\infty {(-1)^n \frac{\Ga(\frac13+\frac n3)}{\Ga(1+\frac n3)} \sin\frac{2\pi n}3}. \tag{10}\label{10} \end{equation*}

We have \begin{equation*} s=\Im S(z) \tag{20}\label{20} \end{equation*} and \begin{equation*} S(z)=\sum_{r=0}^2 S_r(z), \tag{30}\label{30} \end{equation*} where \begin{equation*} S_r(z):=\sum_{k=0}^\infty a_{3k+r}z^{3k+r},\quad a_n:=\frac{\Ga(\frac13+\frac n3)}{\Ga(1+\frac n3)}, \end{equation*} $z:=-e^{2\pi i/3}$. Next, for $k=0,1,\dots$ and $r=0,1,2$, \begin{equation*} a_{3k+r}=\frac{\Ga(\frac{1+r}3+k)}{\Ga(1+\frac r3+k)} =\frac{\Ga(\frac{1+r}3)}{\Ga(1+\frac r3)}\frac{(\frac{1+r}3)_k}{(1+\frac r3)_k}, \end{equation*} where $(x)_k:=\prod_{j=0}^{k-1}(x+j)$. So, by the series definition of the hypergeometric function, for $r=0,1,2$, \begin{equation*} S_r(z)=\frac{\Ga(\frac{1+r}3)}{\Ga(1+\frac r3)}\; _2F_1\Big(\frac{1+r}3,1;1+\frac r3;z^3\Big)z^r. \tag{40}\label{40} \end{equation*}

Collecting \eqref{10}, \eqref{20}, \eqref{30}, and \eqref{40}, we get the following expression of the sum of the series in question in terms of the gamma and hypergeometric functions: $$\sum_{n=1}^\infty {(-1)^n \frac{\Ga(\frac13+\frac n3)}{\Ga(1+\frac n3)} \sin\frac{2\pi n}3} \\ =\Im\sum_{r=0}^2 \frac{\Ga(\frac{1+r}3)}{\Ga(1+\frac r3)}\; _2F_1\Big(\frac{1+r}3,1;1+\frac r3;z^3\Big)z^r, $$ with $z=-e^{2\pi i/3}$.

Answer 6

A Couple of Preliminary Integrals $$ \begin{align} &\int_0^1\frac1{1+t}(1-t)^{-\frac13}\,\mathrm{d}t\\ &=\int_0^1\frac1{2-t}t^{-\frac13}\,\mathrm{d}t\tag{1a}\\ &=\frac1{2^{1/3}}\int_0^{2^{-\frac13}}\frac{3t}{1-t^3}\,\mathrm{d}t\tag{1b}\\ &=\frac1{2^{1/3}}\int_0^{2^{-\frac13}}\left(\frac1{1-t}-\frac{1-t}{1+t+t^2}\right)\,\mathrm{d}t\tag{1c}\\ &=-\frac1{2^{1/3}}\log\left(1-2^{-1/3}\right)-\frac1{2^{1/3}}\int_0^{2^{-\frac13}}\frac{\sqrt3-\frac2{\sqrt3}\left(t+\frac12\right)}{1+\frac43\left(t+\frac12\right)^2}\frac2{\sqrt3}\,\mathrm{d}t\tag{1d}\\ &=-\frac1{2^{1/3}}\log\left(1-2^{-1/3}\right)-\frac1{2^{1/3}}\int_{\frac1{\sqrt3}}^{\frac{1+2^{2/3}}{\sqrt3}}\frac{\sqrt3-u}{1+u^2}\,\mathrm{d}u\tag{1e}\\ &=-\frac1{2^{1/3}}\log\left(1-2^{-1/3}\right)-\frac{\sqrt3}{2^{1/3}}\tan^{-1}\left(\frac{\sqrt3}{1+2^{4/3}}\right)\\ &\phantom{={}}+\frac1{2^{4/3}}\log\left(1+2^{-1/3}+2^{-2/3}\right)\tag{1f}\\ &=-\frac3{2^{4/3}}\log\left(2^{1/3}-1\right)-\frac{\sqrt3}{2^{1/3}}\tan^{-1}\left(\frac{\sqrt3}{1+2^{4/3}}\right)\tag{1g} \end{align} $$ Explanation:
$\text{(1a):}$ substitute $t\mapsto1-t$
$\text{(1b):}$ substitute $t\mapsto2t^3$
$\text{(1c):}$ partial fractions
$\text{(1d):}$ integrate the first fraction
$\phantom{\text{(1d):}}$ collect terms in the second fraction
$\text{(1e):}$ $u=\frac2{\sqrt3}\left(t+\frac12\right)$
$\text{(1f):}$ integrate the arctan and logarithm
$\text{(1g):}$ simplify logs $$ \begin{align} \int_0^1\frac{t^{-\frac13}}{1+t}(1-t)^{-\frac13}\,\mathrm{d}t &=\int_0^\infty\left(\frac{t}{(1+t)^2}\right)^{-1/3}\frac{1+t}{1+2t}\frac{\mathrm{d}t}{(1+t)^2}\tag{2a}\\ &=\int_0^\infty(t(1+t))^{-1/3}\frac{\mathrm{d}t}{1+2t}\tag{2b}\\ &=\frac{2\pi i}{2-2e^{-2\pi i/3}}\frac{e^{-\pi i/3}}{2^{1/3}}\tag{2c}\\ &=\frac\pi{2^{1/3}\sqrt3}\tag{2d} \end{align} $$ Explanation:
$\text{(2a):}$ substitute $t\mapsto\frac{t}{1+t}$
$\text{(2b):}$ simplify
$\text{(2c):}$ contour integration says
$\phantom{\text{(2c):}}$ $\left(2-2e^{-2\pi i/3}\right)\int_0^\infty(t(1+t))^{-1/3}\frac{\mathrm{d}t}{1+2t}=2\pi i\frac{e^{-\pi i/3}}{2^{1/3}}$
$\text{(2d):}$ simplify

Contour Integration for $\bf{(2c)}$

We are integrating $\frac{(z(1+z))^{-1/3}}{1+2z}$ over the contour. The integral over the semi-circular pieces vanishes as the outer radius goes to infinity and the inner radius goes to $0$.

The integral over the straight line just above the right branch cut is the integral in question. The values along the straight line under the right branch-cut are $e^{-2\pi i/3}$ times the values just above, but the direction is reversed, so it contributes $-e^{-2\pi i/3}$ times the integral.

The circle can be rotated about the point $-1/2$ by $z\mapsto-1-z$. This map leaves the integrals as they were. Thus, the integral over the straight lines is $2-2e^{-2\pi i/3}$ times the integral in question.

The residue of $\frac{(z(1+z))^{-1/3}}{1+2z}$ at $z=-1/2$ is $\frac{e^{-\pi i/3}}{2^{1/3}}$, so the integral over the contour is $2\pi i\frac{e^{-\pi i/3}}{2^{1/3}}$

Putting the Pieces Together $$ \begin{align} &\sum_{n=1}^\infty(-1)^n\frac{\Gamma\!\left(\frac13+\frac n3\right)}{\Gamma\!\left(1+\frac n3\right)}\sin\left(\frac{2\pi n}3\right)\\ &=-\sum_{n=1}^\infty\sin\left(\frac{\pi n}3\right)\frac{\Gamma\!\left(\frac13+\frac n3\right)}{\Gamma\!\left(1+\frac n3\right)}\tag{3a}\\ &=-\frac1{\Gamma\!\left(\frac23\right)}\sum_{n=1}^\infty\sin\left(\frac{\pi n}3\right)\int_0^1t^{\frac n3-\frac23}(1-t)^{-\frac13}\,\mathrm{d}t\tag{3b}\\ &=-\frac{\sqrt3}{2\Gamma\!\left(\frac23\right)}\int_0^1\frac{1+t^{-\frac13}}{1+t}(1-t)^{-\frac13}\,\mathrm{d}t\tag{3c}\\ &=-\frac{\sqrt3}{2\Gamma\!\left(\frac23\right)}\left(-\frac3{2^{4/3}}\log\left(2^{1/3}-1\right)-\frac{\sqrt3}{2^{1/3}}\tan^{-1}\left(\frac{\sqrt3}{1+2^{4/3}}\right)+\frac\pi{2^{1/3}\sqrt3}\right)\tag{3d}\\ &=\frac3{2^{7/3}\Gamma\!\left(\frac23\right)}\left(\sqrt3\log\left(2^{1/3}-1\right)-2\tan^{-1}\left(\frac{\sqrt3}{1+2^{2/3}}\right)\right)\tag{3e}\\[6pt] &\approx-1.544563278574849653\tag{3f} \end{align} $$ Explanation:
$\text{(3a):}$ $(-1)^n\sin\left(\frac{2\pi n}3\right)=-\sin\left(\frac{\pi n}3\right)$
$\text{(3b):}$ $\frac{\Gamma\left(\frac13+\frac n3\right)}{\Gamma\left(1+\frac n3\right)}=\frac1{\Gamma\left(\frac23\right)}\int_0^1t^{\frac n3-\frac23}(1-t)^{-\frac13}\,\mathrm{d}t$
$\text{(3c):}$ $\sum\limits_{n=1}^\infty\sin\left(\frac{\pi n}3\right)t^{\frac n3}=\frac{\sqrt3}2\frac{t^{\frac13}+t^{\frac23}}{1+t}$
$\text{(3d):}$ apply $(1)$ and $(2)$
$\text{(3e):}$ $\frac\pi3-\arctan\left(\frac{\sqrt3}{1+2^{4/3}}\right)=\arctan\left(\frac{\sqrt3}{1+2^{2/3}}\right)$
$\text{(3f):}$ evaluate