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Question Number 136519 by Dwaipayan Shikari last updated on 22/Mar/21

1 - {(\frac{1}{2})}^{2 k} + {(\frac{1.3}{2.4})}^{2 k} - {(\frac{1.3.5}{2.4.6})}^{2 k} + {(\frac{1.3.5.7}{2.4.6.8})}^{2 k} - \dots .

F i n d t h e g e n e r a l f o r m

Answered by mindispower last updated on 22/Mar/21

1 + \sum_{n ⩾ 1} \frac{{(- 1)}^{n} . \underset{m = 0}{\prod^{n - 1}} {(2 m + 1)}^{2 k}}{\underset{m = 0}{\prod^{n - 1}} . {(2 (m + 1))}^{2 k}}

= 1 + \sum_{n ⩾ 1} . {(- 1)}^{n} \frac{\underset{m = 0}{\prod^{n - 1}} {(m + \frac{1}{2})}^{2 k}}{\underset{m = 0}{\prod^{m = n - 1}} {(m + 1)}^{2 k}}

= 1 + \sum_{n ⩾ 1} \frac{{(\prod_{m ⩽ n - 1} (\frac{1}{2} + m))}^{2 k}}{{(\prod_{m ⩽ n - 1} (1 + m))}^{2 k - 1}} . \frac{{(- 1)}^{n}}{n!}

= 1 + \sum_{n ⩾ 1} . \frac{{(\frac{1}{2})}_{n}^{2 k} .}{{(1)}^{2 k - 1}} . \frac{{(- 1)}^{n}}{n!}

=_{2 k} F_{2 k - 1} | \begin{matrix} (\frac{1}{2}), \dots \dots \dots (\frac{1}{2}) \\ ; (1), (1) \dots \dots . (1); [- 1] \end{matrix} |

Commented by Dwaipayan Shikari last updated on 22/Mar/21

T h a n k s s i r!

Commented by mindispower last updated on 23/Mar/21

p l e a s u r