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.

#Calculus #DoubleSum #InfiniteSeries #DoubleFactorial

Easy Math Editor

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Comments

CARNAGE D:

Jake Lai - 6 years, 2 months ago

Wallis' is everything I can see!

Kartik Sharma - 6 years, 2 months ago

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.

Hasan Kassim - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$