"Author introduces an alternating series:∑n=1∞(−1)n(2n)!4n(n!)2" was answered by students like you

Layla-Rose Ellison

Answered question

2022-02-22

Author introduces an alternating series:

\sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{(2 n)!}{4^{n} {(n!)}^{2}}

toeneepnla

Beginner2022-02-23Added 4 answers

We are looking at

\sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{(2 n)!}{4^{n} {(n!)}^{2}}

a_{n} = \frac{(2 n)!}{4^{n} {(n!)}^{2}}

\frac{a_{n + 1}}{a_{n}} = \frac{\frac{(2 (n + 1))!}{4^{n + 1} {((n + 1)!)}^{2}}}{\frac{(2 n)!}{4^{n} {(n!)}^{2}}}

= \frac{(2 (n + 1))! 4^{n} {(n!)}^{2}}{4^{n + 1} {((n + 1)!)}^{2} (2 n)!}

= \frac{(2 n + 1) (2 n + 2)}{4 {(n + 1)}^{2}}

= \frac{2 n + 1}{2 (n + 1)}

= 1 - \frac{1}{2 (n + 1)}

a_{n}

is strictly decreasing. Since

\sum \frac{1}{2 (n + 1)}

diverges

a_{n} \to 0

. By the alternating series test,

\sum {(- 1)}^{n} a_{n}

converges
Also,

\frac{a_{n + 1}}{a_{n}} = 1 - \frac{1}{2 (n + 1)} > 1 - \frac{1}{n + 1} = \frac{n}{n + 1}

\frac{a_{N}}{a_{M}} = \prod_{n = M}^{N - 1} \frac{a_{n + 1}}{a_{n}} < \prod_{j = M}^{N - 1} \frac{n}{n + 1} = \frac{M}{N}

. Therefore, since

\sum \frac{1}{N}

diverges,

\sum a_{n}

Abbey Hope

Beginner2022-02-24Added 7 answers

Its

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