Verifying $ \sum_{k=0}^{\infty}{3k \choose k}\frac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\frac{2}{27}\right)^k=\frac{1}{4}$

Question

$$ \sum_{k=0}^{\infty}{3k \choose k}\dfrac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\dfrac{2}{27}\right)^k=\dfrac{1}{4} $$ After some simplification, I got the following result: $$ \sum_{k=0}^{\infty}\left\{{3k \choose k}+\dfrac{27}{2k}{3k-4 \choose 2k-3}\right\}\left(\dfrac{2}{27}\right)^k. $$ Now, how I can proceed?

Answer 1

Starting with $$ \eqalign{ & S = \sum\limits_{k = 0}^\infty { \binom{3k}{k}{{9k^2 - 3k - 1} \over {\left( {3k - 1} \right)\left( {3k - 2} \right)}}\left( {{2 \over {27}}} \right)^{\,k} } = \cr & = \sum\limits_{k = 0}^\infty { \binom{3k}{k} \left( {1 + {1 \over {\left( {3k - 1} \right)}} + {1 \over {\left( {3k - 2} \right)}}} \right)\left( {{2 \over {27}}} \right)^{\,k} } \cr} $$ we have that $$ \binom{3k}{k} = {{\Gamma \left( {3k + 1} \right)} \over {\Gamma \left( {2k + 1} \right)}}{1 \over {k!}} $$

Using the $n$-plication formula for the Gamma function $$ \Gamma \left( {n\,z + 1} \right) = \Gamma \left( {n\,\left( {z + 1/n} \right)} \right)\quad = {{n^{\,n\,z + 1/2} } \over {\left( {2\,\pi } \right)^{\left( {n - 1} \right)/2} }}\prod\limits_{1\, \le \,j\, \le \,n} {\Gamma \left( {z + {j \over n}} \right)} $$ we get $$ \eqalign{ & {{\Gamma \left( {3\,k + 1} \right)} \over {\Gamma \left( {2\,k + 1} \right)}}\quad = {{3^{\,3\,k + 1/2} } \over {2^{\,2\,k + 1/2} \sqrt {2\,\pi } ^\, }}{{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right) \Gamma \left( {k + 1} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)\Gamma \left( {k + 1} \right)}} = \cr & = \sqrt {{3 \over {4\pi }}} {{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)}}\left( {{27 \over 4}} \right)^{\,k} \cr & {{\Gamma \left( {3\,k + 1} \right)} \over {\Gamma \left( {2\,k + 1} \right)\left( {3k - 1} \right)}} = \sqrt {{3 \over {4\pi }}} {{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right)} \over {3\Gamma \left( {k + {1 \over 2}} \right)\left( {k - 1/3} \right)}}\left( {{27 \over 4}} \right)^{\,k} = \cr & = \sqrt {{3 \over {4\pi }}} {1 \over 3}{{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k - {1 \over 3}} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)}}\left( {{27 \over 4}} \right)^{\,k} \cr & {{\Gamma \left( {3\,k + 1} \right)} \over {\Gamma \left( {2\,k + 1} \right)\left( {3k - 2} \right)}} = \sqrt {{3 \over {4\pi }}} {{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right)} \over {3\Gamma \left( {k + {1 \over 2}} \right)\left( {k - 2/3} \right)}}\left( {{27 \over 4}} \right)^{\,k} = \cr & = \sqrt {{3 \over {4\pi }}} {1 \over 3}{{\Gamma \left( {k - {2 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)}}\left( {{27 \over 4}} \right)^{\,k} \cr} $$

So $$ \eqalign{ & S = \cr & = \sum\limits_{k = 0}^\infty {\binom{3k}{k} \left( {1 + {1 \over {\left( {3k - 1} \right)}} + {1 \over {\left( {3k - 2} \right)}}} \right)\left( {{2 \over {27}}} \right)^{\,k} } = \cr & = {1 \over 2}\sqrt {{1 \over {3\pi }}} \sum\limits_{k = 0}^\infty { \left( {3{{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)}} + {{\Gamma \left( {k + {1 \over 3}} \right)\Gamma \left( {k - {1 \over 3}} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)}} + {{\Gamma \left( {k - {2 \over 3}} \right)\Gamma \left( {k + {2 \over 3}} \right)} \over {\Gamma \left( {k + {1 \over 2}} \right)}}} \right){{\left( {1/2} \right)^{\,k} } \over {k!}}} \cr} $$

Using the Hypergeometric function that becomes $$ \eqalign{ & S = \cr & = {1 \over 2}\sqrt {{1 \over {3\pi }}} \left( \matrix{ 3{{\Gamma \left( {1/3} \right)\Gamma \left( {2/3} \right)} \over {\Gamma \left( {1/2} \right)}}{}_2F_1 \left( {\left. {\matrix{ {1/3,\;2/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) + \hfill \cr + {{\Gamma \left( {1/3} \right)\Gamma \left( { - 1/3} \right)} \over {\Gamma \left( {1/2} \right)}}{}_2F_1 \left( {\left. {\matrix{ {1/3,\; - 1/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) + \hfill \cr + {{\Gamma \left( {2/3} \right)\Gamma \left( { - 2/3} \right)} \over {\Gamma \left( {1/2} \right)}}{}_2F_1 \left( {\left. {\matrix{ {2/3,\; - 2/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) \hfill \cr} \right) = \cr & = {}_2F_1 \left( {\left. {\matrix{ {1/3,\;2/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) - {}_2F_1 \left( {\left. {\matrix{ {1/3,\; - 1/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) - {1 \over 2}{}_2F_1 \left( {\left. {\matrix{ {2/3,\; - 2/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) \cr} $$

Since the hypergeometric for the variable $z=1/2$ follows the following formulas (see e.g. this link) $$ \eqalign{ & {}_2F_1 \left( {\left. {\matrix{ {a,\;b} \cr {{{a + b} \over 2}} \cr } \,} \right|\;{1 \over 2}} \right) = \sqrt \pi \;\Gamma \left( {{{a + b} \over 2}} \right)\left( {{1 \over {\Gamma \left( {{{a + 1} \over 2}} \right)\Gamma \left( {{b \over 2}} \right)}} + {1 \over {\Gamma \left( {{a \over 2}} \right)\Gamma \left( {{{b + 1} \over 2}} \right)}}} \right) \cr & {}_2F_1 \left( {\left. {\matrix{ {a,\;b} \cr {{{a + b + 1} \over 2}} \cr } \,} \right|\;{1 \over 2}} \right) = \sqrt \pi \;\Gamma \left( {{{a + b + 1} \over 2}} \right){1 \over {\Gamma \left( {{{a + 1} \over 2}} \right)\Gamma \left( {{{b + 1} \over 2}} \right)}} \cr} $$ then $$ \eqalign{ & {}_2F_1 \left( {\left. {\matrix{ {1/3,\;2/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) = \pi \left( {{1 \over {\Gamma \left( {2/3} \right)\Gamma \left( {1/3} \right)}} + {1 \over {\Gamma \left( {1/6} \right)\Gamma \left( {1 - 1/6} \right)}}} \right) = \cr & = \left( {{3 \over {2\sqrt 3 }} + {1 \over 2}} \right) = {{\sqrt 3 + 1} \over 2} \cr & {}_2F_1 \left( {\left. {\matrix{ {1/3,\; - 1/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) = \pi {1 \over {\Gamma \left( {2/3} \right)\Gamma \left( {1/3} \right)}} = {{\sqrt 3 } \over 2} \cr & {}_2F_1 \left( {\left. {\matrix{ {2/3,\; - 2/3} \cr {1/2} \cr } \,} \right|\;{1 \over 2}} \right) = \pi {1 \over {\Gamma \left( {1 - {1 \over 6}} \right)\Gamma \left( {{1 \over 6}} \right)}} = {1 \over 2} \cr} $$

And $S= 1/4$ follows.

Answer 2

By Lagrange's inversion theorem (see Bring radical for the quintic analogue) $$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{4}{27}\right)^k x^{2k} = \frac{\cos\left(\frac{1}{3}\arcsin x\right)}{\sqrt{1-x^2}} \tag{1}$$ and by partial fraction decomposition $$ \frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+\frac{1}{3k-1}+\frac{1}{3k-2}\tag{2} $$ hence by evaluating $(1)$ at $x=\frac{1}{\sqrt{2}}$ we immediately get $$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{2}{27}\right)^k = \frac{1+\sqrt{3}}{2}. \tag{3}$$ By reindexing the original series equals $$-\frac{1}{2}+\sum_{k\geq 0}\binom{3k}{k}\frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}\left[1+\frac{1}{3k+1}+\frac{1}{3k+2}\right]\left(\frac{2}{27}\right)^k \tag{4}$$ hence it is enough to compute the integrals over $(0,1/\sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $\frac{3}{2\sqrt{2}}(\sqrt{3}-1)$ and $\frac{3}{16}(5-2\sqrt{3})$ by the substitution $x\mapsto\sin\theta$. By re-combining these pieces the claim is proved.