Solved: "Prove that∑n=0∞(2n+1)!23n(n!)2converges to 22"

lovesickgirlsrd5

Answered question

2022-02-24

Prove that

\sum_{n = 0}^{\infty} \frac{(2 n + 1)!}{2^{3 n} {(n!)}^{2}}

converges to

2 \sqrt{2}

Asa Buck

Beginner2022-02-25Added 8 answers

With

| μ | < \frac{1}{4}

\sum_{n = 0}^{\infty} μ^{n} \frac{(2 n + 1)!}{{(n!)}^{2}} = \sum_{n = 0}^{\infty} (2 n + 1) μ^{n} (\begin{matrix} 2 n \\ n \end{matrix}) = (2 μ \frac{d}{d μ} + 1) \sum_{n = 0}^{\infty} μ^{n} (\begin{matrix} 2 n \\ n \end{matrix})

\sum_{n = 0}^{\infty} μ^{n} (\begin{matrix} 2 n \\ n \end{matrix}) = \sum_{n = 0}^{\infty} μ^{n} \oint_{| z | = 1} \frac{{(1 + z)}^{2 n}}{z^{n + 1}} \frac{d z}{2 π i} = \oint_{| z | = 1} \frac{1}{z} \sum_{n = 0}^{\infty} {[\frac{{(1 + z)}^{2} μ}{z}]}^{n} \frac{d z}{2 π i}

= \oint_{| z | = 1} \frac{1}{z} \frac{1}{1 - {(1 + z)}^{2} \frac{μ}{z}} \frac{d z}{2 π i} = - \oint_{| z | = 1} \frac{1}{μ z^{2} + (2 μ - 1) z + μ} \frac{d z}{2 π i}

= - \oint_{| z | = 1} \frac{1}{μ (z - r_{-}) (z - r_{+})} \frac{d z}{2 π i}

where

r_{\pm} = \frac{1 - 2 μ \pm \sqrt{1 - 4 μ}}{2 μ}

Note that

| r_{-} | < 1

and

| r_{+} | > 1

such that:

\sum_{n = 0}^{\infty} μ^{n} (\begin{matrix} 2 n \\ n \end{matrix}) <

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