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Question Number 124347 by Dwaipayan Shikari last updated on 02/Dec/20

Commented by Dwaipayan Shikari last updated on 02/Dec/20

$W h a t w i l l b e t h e a p p r o a c h t o t h i s p r o b l e m ?$
	Answered by mindispower last updated on 03/Dec/20

	$= 1 + \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{4 k + 1} . \underset{i = 0}{\prod^{k - 1}} \frac{{(2 + 3 i)}^{3}}{{(3 + 3 i)}^{3}}$ $= 1 + \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{4 k + 1} . \underset{i = 0}{\prod^{i = k - 1}} \frac{{(\frac{2}{3} + i)}^{3}}{{(1 + i)}^{3}}$ $= 1 + \sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k!} . \frac{{(\frac{2}{3})}_{k} . {(\frac{2}{3})}_{k} {(\frac{2}{3})}_{k}}{{(1)}_{k} {(1)}_{k}}$ $\frac{1}{4 k + 1} = \frac{(\frac{1}{4}) . . . . (\frac{1}{4} + k - 1)}{(\frac{5}{4}) . . . . (\frac{5}{4} + k - 1)} = \frac{{(\frac{1}{4})}_{k}}{{(\frac{5}{4})}_{k}}$ $S = 1 + \sum_{k ⩾ 1} \frac{{(\frac{2}{3})}_{k} {(\frac{2}{3})}_{k} {(\frac{2}{3})}_{k} {(\frac{1}{4})}_{k}}{{(1)}_{k} {(1)}_{k} {(\frac{5}{4})}_{k}} . \frac{{(- 1)}^{k}}{k!}$ $S =_{4} F_{3} (\frac{2}{3}, \frac{2}{3}, \frac{2}{3}, \frac{1}{4}; 1, 1, \frac{5}{4}, - 1)$
	Commented by Dwaipayan Shikari last updated on 03/Dec/20

	$I n a c l o s e d f o r m o f G a m m a f u n c t i o n ?$
	Commented by mindispower last updated on 03/Dec/20

	$y e s i t h i n k_{p} F_{q} t h e r i s r e l a t i o n b e t e w e e n$ $i ? w i l l t r y$
	Commented by Dwaipayan Shikari last updated on 03/Dec/20

	$T h a n k i n g y o u$
	Commented by Dwaipayan Shikari last updated on 03/Dec/20

	$T h e a n s w e r i s g i v e n a s \frac{3 \sqrt{3}}{4 {(2 π)}^{5}} Γ^{9} (\frac{1}{3})$
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