Resolved: "Why is:1m+1=∑n=0∞2−2n−m−1(2n+m)!(n+m+1)!n!"

embeucadaoo8

Answered question

2022-02-25

Why is:

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

razlikaml42

Beginner2022-02-26Added 5 answers

A posible idea could be to use

f (x) = \sum_{n = 0}^{\infty} \frac{2^{- (2 n + m + 1)} (2 n + m)!}{(n + m + 1)! n!} x^{- n} = \frac{2 F_{1} (\frac{m + 1}{2}, \frac{m + 2}{2}; m + 2; \frac{1}{x})}{(m + 1) 2^{m + 1}}

which is true even if m is not an integer.
Now, working a litle the expression

f (x) = \frac{x}{m + 1} {(1 + \sqrt{\frac{x - 1}{x}})}^{- m} (1 - \sqrt{\frac{x - 1}{x}})

Gene Espinosa

Beginner2022-02-27Added 7 answers

I guess I post my approch which is based on the idea of Claude Leibovici for clarity.
Using Legendre duplication we have

(2 n + m) \neq Γ (2 n + m + 1) = Γ (n + \frac{m + 1}{2}) Γ (n + \frac{m + 2}{2}) \frac{2^{2 n + m}}{\sqrt{π}}

So we have

\sum_{n = 0}^{\infty} \frac{2^{- 2 n - m - 1} (2 n + m)!}{(n + m + 1)! n!} = \frac{1}{2 \sqrt{π}} \sum_{n = 0}^{\infty} \frac{Γ (n + \frac{m + 1}{2}) Γ (n + \frac{m + 2}{2})}{Γ (n + m + 2) n!}

= \frac{1}{2 \sqrt{π}} \frac{Γ (\frac{m + 1}{2}) Γ (\frac{m + 2}{2})}{Γ (m + 2)} F (\frac{m + 1}{2}, \frac{m + 2}{2}, m + 2, 1)

= \frac{1}{2 \sqrt{π}} \frac{Γ (\frac{m + 1}{2}) Γ (\frac{m + 2}{2})}{Γ (m + 2)} \frac{Γ (m + 2) Γ (m + 2 - \frac{m + 1}{2} - \frac{m + 2}{2})}{Γ (m + 2 - \frac{m + 1}{2}) Γ (m + 2 - \frac{m + 2}{2})}

= \frac{1}{2 \sqrt{π}} \frac{Γ (\frac{m + 1}{2}) Γ (\frac{m + 2}{2})}{Γ (m + 2)} F (\frac{m + 1}{2}, \frac{m + 2}{2}, m + 2, 1)

= \frac{1}{2 \sqrt{π}} \frac{Γ (\frac{m + 1}{2}) Γ (\frac{m + 2}{2})}{}

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