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After Playing With Integrals!

$\displaystyle {\color{grey}{\sum_{n=0}^{\infty} \sum_{i=0}^n \frac{(2n)!!(2i+1)!!}{(2i)!!(2n+1)!! } (-b)^{i+n} = \frac{1}{(1+b)^2}\Bigg(\frac{1}{2\sqrt{b}} \ln \Big(\frac{1+\sqrt{b}}{1-\sqrt{b}} \Big) + \frac{b}{1-b} \Bigg)}}$

Prove the above equality for $$0<b<1$$.

Note: I got this fact after manipulating some integrals (i.e. proved it indirectly). I wonder how brilliant members would solve it ! Share your thoughts and hints if you don't have the complete answer.



This note is a part of the set Sequences and Series Challenges.

Note by Hasan Kassim
2 years, 11 months ago

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CARNAGE D:

- 2 years, 11 months ago

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Wallis' is everything I can see!

- 2 years, 11 months ago

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That's right, Wallis' is a way to start. But as you will see, it would be messy.I wonder if a simple solution can be found.

- 2 years, 11 months ago

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