it-seems-too-hard-for-many-here-to-post-their-questions-as-questions-answers-as-answers-and-comments-as-comments-hereby-I-introduce-the-next-step-I-ll-post-answers-and-the-task-is-find-questions

Question Number 110539 by Her_Majesty last updated on 29/Aug/20

i t s e e m s t o o h a r d f o r m a n y h e r e t o p o s t

t h e i r q u e s t i o n s a s q u e s t i o n s, a n s w e r s a s

a n s w e r s a n d c o m m e n t s a s c o m m e n t s \dots

h e r e b y I i n t r o d u c e t h e n e x t s t e p : I^{'} l l p o s t

a n s w e r s a n d t h e t a s k i s, f i n d q u e s t i o n s

t o t h e s e a n s w e r s

(1) ζ (3) + π l n 2

(2) π H_{0} (7)

(3) t r u e \forall x \in C ∖ Q

(4)_{2} F_{1} (\frac{3}{2}, \frac{1}{5}, \frac{2}{3}, {s i n}^{- 1} \frac{x + 1}{x - 1})

Commented by Eric002 last updated on 29/Aug/20

i t h i n k o n l y p r o f . m a t h m i n d a n d

a n d p r o f . * M * t h + e t + s h a v e k n o w l e d g e

o f (4) h y p e r g e o m e t r i c f u n c t i o n a n d t h e y

{h a v e n}^{'} t b e e n a r o u n d f o r a l o n g t i m e

Commented by Dwaipayan Shikari last updated on 29/Aug/20

1) \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} + ((\frac{22}{7} - \int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}}) \underset{n = 1}{\sum^{\infty}} \frac{1}{n 2^{n}})

O r \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} + ((\frac{22}{7} - \int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}}) \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1}}{n})

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} + \frac{22}{7} \underset{n = 1}{\sum^{\infty}} \frac{1}{n 2^{n}} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n 2^{n}} (\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}})

Commented by Dwaipayan Shikari last updated on 29/Aug/20

Infinitely many questions ��

Commented by Her_Majesty last updated on 29/Aug/20

��