We have an answer to "Simplify the power series∑n=0∞(2n)!n!(n+1)!⋅pn(1−p)n"

Brittney Cuevas

Answered question

2022-02-25

Simplify the power series

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

vefibiongedogn7z

Beginner2022-02-26Added 6 answers

Let

x = p (1 - p)

. We have

P (p) = \sum_{n = 0}^{\infty} \frac{(2 n)!}{n! (n + 1)!} p^{n} {(1 - p)}^{n} = \sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{x^{n}}{n + 1}

Note the standard power series

\sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) x^{n} = \frac{1}{\sqrt{1 - 4 x}}

which is convergent for

| x | < \frac{1}{4}

, with antiderivative

\sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{x^{n + 1}}{n + 1} = \frac{1}{2} (1 - \sqrt{1 - 4 x})

Therefore, if

p \neq \frac{1}{2}

(so that

x < \frac{1}{4}

P (p) = \frac{1}{2 p (1 - p)} (1 - \sqrt{1 - 4 p (1 - p)}) = \frac{1 - \sqrt{{(1 - 2 p)}^{2}}}{2 p (1 - p)} = \frac{1 - | 1 - 2 p |}{2 p (1 - p)}

This identity is still valid for

p = \frac{1}{2} (x = \frac{1}{4})

) by Abel's radial convergence theorem (or more simply, monotone convergence).

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