I want to know how to compute the closed form for the series∑n=0∞(2n)!(n!)2(|A|2)2n,|A|&lt;1,

Applying the Taylor development to the function f(A)=11−A2f(A)=(1−A2)−12=1+11!(−12)(−A2)+12!(−12)(−12−1)(−A2)2+…+1n!(∏i=0n−1(−12−i))(−A2)n+…=1+∑n=1+∞(1n!(−A2)n∏i=0n−1(−12−i))=1+∑n=1+∞(2n−1)!!2nn!(A2)n=1+∑n=1+∞((2n−1)!!2nn!×(2n)!!2nn!)(A2)n=1+∑n=1+∞(2n)!(2n)2(n!)2(A2)n=∑n=0+∞(2n)!(n!)2(A22)nQ.E.D

Expert Answer to "I want to know how to compute the..."

OIPLZLqm6

Answered question

2022-02-26

I want to know how to compute the closed form for the series

\sum_{n = 0}^{\infty} \frac{(2 n)!}{{(n!)}^{2}} {(\frac{| A |}{2})}^{2 n}, | A | < 1,

Answer & Explanation

Pregazzix2a

Beginner2022-02-27Added 9 answers

Applying the Taylor development to the function

f (A) = \frac{1}{\sqrt{1 - A^{2}}}

f (A) = {(1 - A^{2})}^{- \frac{1}{2}}

= 1 + \frac{1}{1!} (- \frac{1}{2}) (- A^{2}) + \frac{1}{2!} (- \frac{1}{2}) (- \frac{1}{2} - 1) {(- A^{2})}^{2} + \dots + \frac{1}{n!} (\prod_{i = 0}^{n - 1} (- \frac{1}{2} - i)) {(- A^{2})}^{n} + \dots

= 1 + \sum_{n = 1}^{+ \infty} (\frac{1}{n!} {(- A^{2})}^{n} \prod_{i = 0}^{n - 1} (- \frac{1}{2} - i))

= 1 + \sum_{n = 1}^{+ \infty} \frac{(2 n - 1)!!}{2^{n} n!} {(A^{2})}^{n}

= 1 + \sum_{n = 1}^{+ \infty} (\frac{(2 n - 1)!!}{2^{n} n!} \times \frac{(2 n)!!}{2^{n} n!}) {(A^{2})}^{n}

= 1 + \sum_{n = 1}^{+ \infty} \frac{(2 n)!}{{(2^{n})}^{2} {(n!)}^{2}} {(A^{2})}^{n}

= \sum_{n = 0}^{+ \infty} \frac{(2 n)!}{{(n!)}^{2}} {(\frac{A^{2}}{2})}^{n}

Q.E.D

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?